In which quadrants do the following points lie:
(10, -3)
Given coordinate (10,-3) lies in Quadrant IV because as shown in the figure that those coordinate who have (+, -) sign lies in IV quadrants, or can say whose x-axis is “+” and y-axis is “–“ lies in IV Quadrant.
In which quadrants do the following points lie:
(-4, -6)
Given coordinate (-4,-6) lies in Quadrant III because as shown in the figure that those points which have a sign like this (-, -) or can say whose both x-axis and y-axis is “–“ lies in III quadrants. So (-4,-6) lies in III Quadrant.
In which quadrants do the following points lie:
(-8, 6)
Given coordinate (-8,6) lies in Quadrant II because as shown in the figure that those points which have a sign like this (-, +) lies in II quadrants. So (-8,6) lies in III Quadrant. Here also x-axis point is “-” and y-axis point is “+” so (-8,6) lies in Quadrant II.
Given coordinate lies, in Quadrant I because as shown in the figure that those coordinate who have (+, +) sign lies in IV quadrants or can say whose x-axis is “+” and y-axis is also “+“ lies in I Quadrant.
In which quadrants do the following points lie:
(3, 0)
Given coordinate is (3,0). This point lies on the x-axis because its y-axis is on origin. Therefore, it lies on the x-axis between the Quadrant I and Quadrant IV.
In which quadrants do the following points lie:
(0, -5)
Given coordinate is (0, -5). This point lies on the y-axis because its x-axis is on origin. Therefore, it lies on the y-axis between the Quadrant III and Quadrant IV.
Plot the following points in a rectangular coordinate system:
(4, 5)
Here is the graph for coordinate (4,5)
Plot the following points in a rectangular coordinate system:
(-2,-7)
Plot the following points in a rectangular coordinate system:
(6,-2)
Plot the following points in a rectangular coordinate system:
(-4, 2)
Plot the following points in a rectangular coordinate system:
(4, 0)
Plot the following points in a rectangular coordinate system:
(0, 3)
Where does the point having y-coordinate -5 lie?
The point having -5 lies on the on the y-axis on the negative side because here x-axis is 0 and when x-axis is 0 then points lies on the y-axis and when y-axis is 0 then point lies on the x-axis.
We can show it on graph with points (0, -5).
If three vertices of a rectangle are (-2, 0), (2, 0), (2, 1) find the coordinates of the fourth vertex.
Here we have three vertices of rectangle say A (-2, 0) B (2, 0) and C (2, 1) so when we start graphing it on the graph as shown in the graph below then after plotting all three vertices you will get something like this.
Therefore, after joining all the vertices with a line segment, we will get our fourth vertex because in rectangle opposites sides are parallel and all the angles are right angle so, by joining all the lines according to properties of the rectangle you will get the fourth vertex. So fourth vertex of a rectangle is (-2, 1).
Draw the triangle whose vertices are (2, 3), (-4, 2) and (3, -1).
It is easy to draw a triangle when it’s all vertices are given. We have to just locate all the given points on the graph and join them with a line as shown in the graph below.
Step 1. Locate all the vertices on the graph.
Step 2. Joins all the vertices with a line and it will form a triangle.
The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find
the vertices of the triangle.
Given that the base of the equilateral triangle is on the y-axis and mid-point of the base is at the origin, so its figure will be like this as shown.
So here, O(0,0) is the midpoint of the base.
An equilateral triangle has all sides equal so if O is the mid-point of the base BC, so B and C are the two vertices of the triangle. Now we have two vertices of the triangle, which is the base of equilateral triangle lying on the y-axis. Now if base in on y-axis then x-axis are as bisector of the base and so our third vertices will be on the x-axis either left or right.
So now in right ∆BOA
Pythagoras Theorem: In a right-angled triangle the square of the biggest side(hypotenuse) equals the sum of the squares of the other two sides(Perpendicular and base).
BO2 + OA2 = AB2 { By Pythagoras theorem}
a2 + OA2 = (2a) 2
OA2 = 4a2- a2
OA2 = 3a2
OA = ±a√3
So vertices of triangle are A(±a√3,0) B(0,a)and C(0,-a).
Let ABCD be a rectangle such that AB = 10 units and BC = 8 units. Taking AB and AD as x and y-axes respectively, find the coordinates of A, B, C and D.
Since AB and AD both have as an endpoint, we can find the coordinate of A by finding the intersection of the two sides.
So the coordinates of A will be where the x-axis and y-axis intersect.
As we know AB lies on the x-axis so the coordinates of B can be found by using the coordinates of A and changing the x-coordinate by the measure of AB.
As we know the opposite side of a rectangle, AD and BC are congruent. Now if we have a measure of BC, we can simply find the y-coordinate of D.
Since AB and AD are the x and y-axes, A is at(0,0), B is at (10,0), C is at (10,8), and D is at (0,8).
ABCD is a square having a length of a side 20 units. Taking the centre of the square as the origin and x and y-axes parallel to AB and AD respectively, find the coordinates of A, B, C and D.
Square ABCD. Center = O(0,0) Origin.
AB = BC = 20 units.
Y-coordinates of AB = = -10
Y-coordinates of AD = = 10
∴the coordinates are :-
A(-10,-10)
B(10,-10)
C(10,10)
D(-10,10)
Find the distance between the following pair of points:
(0, 0), (- 5, 12)
Given points are (0, 0) and (- 5, 12)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒ S = √169
⇒ S = 13
∴ The distance between the points (0, 0) and (- 5, 12) is 13 units.
Find the distance between the following pair of points:
(4, 5), (- 3, 2)
Given points are (4, 5) and (- 3, 2)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒ S = √58
∴ The distance between the points (4, 5) and (- 3, 2) is √58 units.
Find the distance between the following pair of points:
(5, - 12), (9, - 9)
Given points are (5, - 12) and (9, - 9)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒ S = √25
⇒ S = 5
∴ The distance between the points (5, - 12) and (9, - 9) is 5 units.
Find the distance between the following pair of points:
(- 3, 4), (3, 0)
Given points are (- 3, 4) and (3, 0)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒ S = √52
⇒
⇒ S = 2√13
∴ The distance between the points (- 3, 4) and (3, 0) is 2√13 units.
Find the distance between the following pair of points:
(2, 3), (4, 1)
Given points are (2, 3) and (4, 1)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒ S = √8
⇒
⇒ S = 2√2
∴ The distance between the points (2, 3) and (4, 1) is 2√2 units.
Find the distance between the following pair of points:
(a, b), (- a, - b)
Given points are (a, b) and (- a, - b)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
∴ The distance between the points (a, b) and (- a, - b) is .
Examine whether the points (1, - 1), (- 5, 7) and (2, 6) are equidistant from the point (- 2, 3)?
Given that we need to show that the points (1, - 1), (- 5, 7) and (2, 6) are equidistant from the point (- 2, 3).
We know that distance between two points (x1, y1) and (x2, y2) is
Let S1 be the distance between the points (1, - 1) and (- 2, 3)
⇒ S1 = 5 ..... (1)
Let S2 be the distance between the points (- 5, 7) and (- 2, 3)
⇒ S2 = 5 ..... (2)
Let S3 be the distance between the points (2, 6) and (- 2, 3)
⇒ S3 = 5 ..... (3)
From (1), (2), and (3) we got S1 = S2 = S3 which tells us that (1, - 1), (5, 7) and (2, 5) are equidistant from (- 2, 3).
Find a if the distance between (a, 2) and (3, 4) is 8.
Given that the distance between the points (a, 2) and (3, 4) is 8.
We need to find the value of a.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒ 82 = (a - 3)2 + (- 2)2
⇒ 64 = (a - 3)2 + 4
⇒ (a - 3)2 = 60
⇒ a – 3 = ± √60
⇒ a = 3 ± √60
∴ The values of a are 3±√60.
A line is of length 10 units and one of its ends is (- 2, 3). If the ordinate of the other end is 9, prove that the absicca of the other end is 6 or - 10.
Given that the line has length of 10 units and one of its ends is (- 2, 3).
It is also given that the ordinate of the other end is 9. Let us assume the other end is (x, 9).
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒
⇒ 10 = (x + 2)2 + (- 6)2
⇒ 100 = (x + 2)2 + 36
⇒ (x + 2)2 = 64
⇒ x + 2 = ±8
⇒ x = - 2 + 8 (or) x = - 2 - 8
⇒ x = 6 or 10
∴ Thus proved.
Find the value of y for which the distance between the points P(2, - 3) and Q(10, y) is 10 units.
Given that the distance between the points P(2, - 3) and Q(10, y) is 10.
We need to find the value of y.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒
⇒ 102 = (- 8)2 + (3 + y)2
⇒ 100 = 64 + (3 + y)2
⇒ (3 + y)2 = 36
⇒ 3 + y = ±6
⇒ y = 3 + 6 (or) y = 3 - 6
⇒ y = 9 (or) y = - 3
∴ The values of y are 9, - 3.
Find the distance between the points:
(at12, 2at1) and (at22, 2at2)
Given points are (at12, 2at1) and (at22, 2at2).
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
∴ The distance between the points (at12, 2at1) and (at22, 2at2) is .
Find the distance between the points:
(a - b, b - a) and (a + b, a + b)
Given points are (a - b, b - a) and (a + b, a + b).
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
∴ The distance between the points (a - b, b - a) and (a + b, a + b) is .
Find the distance between the points:
(cosθ, sinθ) and (sinθ, cosθ)
Given points are (cosθ, sinθ) and (sinθ, cosθ).
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
∴ The distance between the points (cosθ, sinθ) and (sinθ, cosθ) is .
Find the point on x - axis which is equidistant from the following pair of points:
(7, 6) and (- 3, 4)
Given points are A(7, 6) and B(- 3, 4).
We need to find a point on x - axis which is equidistant from these points.
Let us assume the point on x - axis be S(x, o).
We know that distance between the points (x1, y1) and (x2, y2) is .
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 7)2 + (0 - 6)2 = (x - (- 3))2 + (0 - 4)2
⇒ (x - 7)2 + (- 6)2 = (x + 3)2 + (- 4)2
⇒ x2 - 14x + 49 + 36 = x2 + 6x + 9 + 16
⇒ 20x = 60
⇒ x = 3
∴ The point on x - axis is (3, 0).
Find the point on x - axis which is equidistant from the following pair of points:
(3, 2) and (- 5, - 2)
Given points are A(3, 2) and B(- 5, - 2).
We need to find a point on x - axis which is equidistant from these points.
Let us assume the point on x - axis be S(x, o).
We know that distance between the points (x1, y1) and (x2, y2) is .
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 3)2 + (0 - 2)2 = (x + 5)2 + (0 - (- 2))2
⇒ (x - 3)2 + (- 2)2 = (x + 5)2 + (2)2
⇒ x2 - 6x + 9 + 4 = x2 + 10x + 25 + 4
⇒ 16x = - 16
⇒ x = - 1
∴ The point on x - axis is (- 1, 0).
Find the point on x - axis which is equidistant from the following pair of points:
(2, - 5) and (- 2, 9)
Given points are A(2, - 5) and B(- 2, 9).
We need to find a point on x - axis which is equidistant from these points.
Let us assume the point on x - axis be S(x, o).
We know that distance between the points (x1, y1) and (x2, y2) is .
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 2)2 + (0 - (- 5))2 = (x - (- 2))2 + (0 - 9)2
⇒ (x - 2)2 + (5)2 = (x + 2)2 + (- 9)2
⇒ x2 - 4x + 4 + 25 = x2 + 4x + 4 + 81
⇒ 8x = - 56
⇒ x = - 7
∴ The point on x - axis is (- 7, 0).
Find the point on y - axis which is equidistant from point (- 5, - 2) and (3, 2).
Given points are A(- 5, - 2) and B(3, 2).
We need to find a point on y - axis which is equidistant from these points.
Let us assume the point on y - axis be S(0, y).
We know that distance between the points (x1, y1) and (x2, y2) is .
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (0 - (- 5))2 + (y - (- 2))2 = (0 - 3)2 + (y - 2)2
⇒ (5)2 + (y + 2)2 = (- 3)2 + (y - 2)2
⇒ 25 + y2 + 4y + 4 = 9 + y2 - 4y + 4
⇒ 8y = - 16
⇒ y = - 2
∴ The point on y - axis is (0, - 2).
Find the point on y - axis which is equidistant from the points A(6, 5) and B(- 4, 3).
Given points are A(6, 5) and B(- 4, 3).
We need to find a point on y - axis which is equidistant from these points.
Let us assume the point on y - axis be S(0, y).
We know that distance between the points (x1, y1) and (x2, y2) is .
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (0 - 6)2 + (y - 5)2 = (0 - (- 4))2 + (y - 3)2
⇒ (- 6)2 + (y - 5)2 = (4)2 + (y - 3)2
⇒ 36 + y2 - 10y + 25 = 16 + y2 - 6y + 9
⇒ 4y = 36
⇒ y = 9
∴ The point on y - axis is (0, 9).
Using distance formula, examine whether the following sets of points are collinear?
(3, 5), (1, 1), (- 2, - 5)
Given points are A(3, 5), B(1, 1) and C(- 2, - 5).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is AC = AB + BC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
⇒ AC = 5√5 ..... (1)
⇒ AB = 2√5 ..... (2)
⇒ BC = 3√5 ..... (3)
From (1), (2), (3) we can see that AB + BC = AC.
∴ The three points are collinear.
Using distance formula, examine whether the following sets of points are collinear?
(5, 1), (1, - 1), (11, 4)
Given points are A(5, 1), B(1, - 1) and C(11, 4).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
⇒ AC = 3√5 ..... (1)
⇒ AB = 2√5 ..... (2)
⇒ BC = 5√5 ..... (3)
From (1), (2), (3) we can see that AB + AC = BC.
∴ The three points are collinear.
Using distance formula, examine whether the following sets of points are collinear?
(0, 0), (9, 6), (3, 2)
Given points are A(0, 0), B(9, 6) and C(3, 2).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
⇒ AC = √13 ..... (1)
⇒ AB = 3√13 ..... (2)
⇒ BC = 2√13 ..... (3)
From (1), (2), (3) we can see that AB = BC + AC.
∴ The three points are collinear.
Using distance formula, examine whether the following sets of points are collinear?
(- 1, 2), (5, 0), (2, 1)
Given points are A(- 1, 2), B(5, 0) and C(2, 1).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
⇒ AC = √10 ..... (1)
⇒ AB = 2√10 ..... (2)
⇒ BC = √10 ..... (3)
From (1), (2), (3) we can see that AC + BC = AB.
∴ The three points are collinear.
Using distance formula, examine whether the following sets of points are collinear?
(1, 5), (2, 3), (- 2, - 11)
Given points are A(1, 5), B(2, 3) and C(- 2, - 11).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is AC = AB + BC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
⇒ AC = √265 ..... (1)
⇒ AB = √5 ..... (2)
⇒ BC = 2√53 ..... (3)
From (1), (2), (3) we can see that we cannot get any linear relationship.
∴ The three points are not collinear.
If A = (6, 1), B = (1, 3) and C = (x, 8), find the value of x such that AB = BC.
Given points are A(6, 1), B(1, 3) and C(x, 8). We need to find the value of x such that AB = BC.
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒ AB = BC
⇒ AB2 = BC2
⇒ (6 - 1)2 + (1 - 3)2 = (1 - x)2 + (3 - 8)2
⇒ (5)2 + (- 2)2 = (1 - x)2 + (- 5)2
⇒ 25 + 4 = 1 - 2x + x2 + 25
⇒ x2 - 2x - 3 = 0
⇒ x2 - 3x + x - 3 = 0
⇒ x(x - 3) + 1(x - 3) = 0
⇒ (x + 1)(x - 3) = 0
⇒ x + 1 = 0 (or) x - 3 = 0
⇒ x = - 1 (or) x = 3
∴ The values of x are - 1 or 3.
Prove that the distance between the points (a + rcosθ, b + rsinθ) and (a, b) is independent of θ.
Given points are A(a + rcosθ, b + rsinθ) and B(a, b).
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒ AB = r
We can see that AB is independent of θ.
∴ Thus proved.
use distance formula to show that the points (cosec2θ, 0), (0, sec2θ) and (1, 1) are collinear.
Given points are A(cosec2θ, 0), B(0, sec2θ ) and C(1, 1).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is AB = AC + BC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
..... (1)
.... - (2)
⇒
⇒
⇒ ..... (3)
Now,
⇒ AC + BC = AB
∴ The three points are collinear.
Using distance formula show that (3, 3) is the centre of the circle passing through the points (6, 2), (0, 4) and (4, 6). Find the radius of the circle.
Given that circle passes through the points A(6, 2), B(0, 4), C(4, 6).
Let us assume O(x, y) be the centre of the circle.
We know that distance from the centre to any point on h circle is equal.
So, OA = OB = OC
We know that distance between two points (x1, y1) and (x2, y2) is
Now,
⇒ OA = OB
⇒ OA2 = OB2
⇒ (x - 6)2 + (y - 2)2 = (x - 0)2 + (y - 4)2
⇒ x2 - 12x + 36 + y2 - 4y + 4 = x2 + y2 - 8y + 16
⇒ 12x - 4y = 24
⇒ 3x - y = 6 ..... (1)
Now,
⇒ OB = OC
⇒ OB2 = OC2
⇒ (x - 0)2 + (y - 4)2 = (x - 4)2 + (y - 6)2
⇒ x2 + y2 - 8y + 16 = x2 - 8x + 16 + y2 - 12y + 36
⇒ 8x + 4y = 36
⇒ 2x + y = 9 .... - (2)
On solving (1) and (2), we get
⇒ x = 3 and y = 3
∴ (3, 3) is the centre of the circle.
We know radius is the distance between the centre and any point on the circle.
Let ‘r’ be the radius of the circle.
⇒ r = √10
∴ The radius of the circle is √10.
If the point (x, y) on the tangent is equidistant from the points (2, 3) and (6, - 1), find the relation between x and y.
Given points are A(2, 3) and B(6, - 1). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is .
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 2)2 + (y - 3)2 = (x - 6)2 + (y - (- 1))2
⇒ (x - 2)2 + (y - 3)2 = (x - 6)2 + (y + 1)2
⇒ x2 - 4x + 4 + y2 - 6y + 9 = x2 - 12x + 36 + y2 + 2y + 1
⇒ 8x - 8y = 24
⇒ x - y = 3
∴ The relation between x and y is x - y = 3.
Find a relation between x and y such that the point (x, y) is equidistant from points (7, 1) and (3, 5).
Given points are A(7, 1) and B(3, 5). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is .
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 7)2 + (y - 1)2 = (x - 3)2 + (y - 5)2
⇒ x2 - 14x + 49 + y2 - 2y + 1 = x2 - 6x + 9 + y2 - 10y + 25
⇒ 8x - 8y = 16
⇒ x - y = 2
∴ The relation between x and y is x - y = 2.
If the distances of P(x, y) from points A(3, 6) and B(- 3, 4) are equal, prove that 3x + y = 5
Given points are A(3, 6) and B(- 3, 4). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is .
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 3)2 + (y - 6)2 = (x - (- 3))2 + (y - 4)2
⇒ (x - 3)2 + (y - 6)2 = (x + 3)2 + (y - 4)2
⇒ x2 - 6x + 9 + y2 - 12y + 36 = x2 + 6x + 9 + y2 - 8y + 16
⇒ 12x + 4y = 20
⇒ 3x + y = 5
∴ Thus proved.
If the point (x, y) be equidistant from the points (a + b, b - a) and (a - b, a + b), prove that
Given points are A(a + b, b - a) and B(a - b, a + b). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is .
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - (a + b))2 + (y - (b - a))2 = (x - (a - b))2 + (y - (a + b))2
⇒ x2 - 2(a + b)x + (a + b)2 + y2 - 2(b - a)y + (b - a)2 = x2 - 2(a - b)x + (a - b)2 + y2 - 2(a + b)y + (a + b)2
⇒ x(- 2a - 2b + 2a - 2b) = y(2b - 2a - 2a - 2b)
⇒ x(- 4b) = y(- 4a)
⇒ x(b) = y(a)
⇒
Applying componendo and dividendo,
⇒
∴ Thus proved.
Prove that the points (3, 4), (8, - 6) and (13, 9) are the vertices of a right angled triangle.
Given points are A(3, 4), B(8, - 6) and C(13, 9).
Let us find the distance between sides AB, BC and CA.
We know that distance between the two points (x1, y1) and (x2, y2) is .
⇒ AB = √125
⇒ BC = √250
⇒ CA = √125
Now,
⇒ AB2 + CA2 = 125 + 125
⇒ AB2 + CA2 = 250
⇒ AB2 + CA2 = BC2
∴ The given points form a right angled isosceles triangle.
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following triangles whose vertices are:
(1, 1), (- ), (- 1, - 1)
Given points are A(1, 1), B(- √3, √3) and C(- 1, - 1).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the two points (x1, y1) and (x2, y2) is
⇒
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We got AB = BC = CA
∴ The given points form an equilateral triangle.
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following triangles whose vertices are:
(0, 2), (7, 0), (2, 5)
Given points are A(0, 2), B(7, 0) and C(2, 5).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is .
⇒
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We got AB≠BC≠CA
∴ The given points form a scalene triangle.
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following triangles whose vertices are:
(- 2, 5), (7, 10), (3, - 4)
Given points are A(- 2, 5), B(7, 10) and C(3, - 4).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is .
⇒
⇒
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⇒
⇒
⇒
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We got AB = CA
Now,
⇒
⇒ AB2 + CA2 = 106 + 106
⇒ AB2 + CA2 = 212
⇒
⇒ AB2 + CA2 = BC2
∴ The given points form a right angles isosceles triangle.
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following triangles whose vertices are:
(4, 4), (3, 5), (- 1, - 1)
Given points are A(4, 4), B(3, 5) and C(- 1, - 1).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is .
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Now,
⇒
⇒ AB2 + CA2 = 2 + 50
⇒ AB2 + CA2 = 52
⇒
⇒ AB2 + CA2 = BC2
∴ The given points form a right - angled triangle.
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following triangles whose vertices are:
(1, 2), (3, 0), (- 1, 0)
Given points are A(1, 2), B(3, 0) and C(- 1, 0).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is .
⇒
⇒
⇒
⇒
⇒ AB = 4
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒ CA = 4
We got AB = BC = CA
∴ The given points form an equilateral triangle.
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following triangles whose vertices are:
(0, 6), (- 5, 3), (3, 1)
Given points are A(0, 6), B(- 5, 3) and C(3, 1).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is .
⇒
⇒
⇒
⇒
⇒
⇒
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⇒
We got AB = CA
Now,
⇒
⇒
⇒ AB2 + CA2 = 68
⇒
⇒ AB2 + CA2 = BC2
∴ The given points form a right - angled isosceles triangle.
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following triangles whose vertices are:
(5, - 2), (6, 4), (7, - 2)
Given points are A(5, - 2), B(6, 4) and C(7, - 2).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is .
⇒
⇒
⇒
⇒
⇒
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We got AB = BC≠CA
∴ The given points form an isosceles triangle.
If A(at2, 2at), B and C(a, 0) be any three points, show that is independent of t.
Given points are A(at2, 2at), B and C(a, 0).
Let us find the distance between sides AB, BC and CA.
We know that distance between the two points (x1, y1) and (x2, y2) is .
⇒ AC = at2 + a
Now,
is independent of t.
If two vertices of an equilateral triangle be (0, 0) and (3, ), find the co - ordinates of the third vertex.
Given that A(0, 0) and B(3, ) are two vertices of an equilateral triangle.
Let us assume C(x, y) be the third vertex of the triangle.
We have AB = BC = CA
We know that the distance between the two points (x1, y1) and (x2, y2) is .
Now,
⇒ BC = CA
⇒ BC2 = CA2
⇒ (3 - x)2 + ( - y)2 = (x - 0)2 + (y - 0)2
⇒ x2 - 6x + 9 + 3 + y2 - 2y = x2 + y2
⇒ 6x = 12 - 2y
..... - (1)
⇒ AB = BC
⇒ AB2 = BC2
⇒ (0 - 3)2 + (0 - )2 = (3 - x)2 + ( - y)2
⇒ 9 + 3 = 9 - 6x + x2 + 3 - 2y + y2
From (1)
⇒
⇒ 48y2 - 48√3y - 288 = 0
⇒ y2–√3y –6 = 0
⇒ y2 - 2√3y + √3y - 6 = 0
⇒ y(y - 2√3) + √3(y - 2√3) = 0
⇒ (y + √3)(y - 2√3) = 0
⇒ y + √3 = 0 (or) y - 2√3 = 0
⇒ y = - √3 (or) y = 2√3
From (1), for y =
⇒ x = 3
From (1), for y = 2
⇒ x = 0
∴ The third vertex of equilateral triangle is (0, 2√3) and (3, √3).
Find the circum - centre and circum - radius of the triangle whose vertices are (- 2, 3), (2, - 1) and (4, 0).
Given that we need to find the circum - centre and circum - radius of the triangle whose vertices are A(- 2, 3), B(2, - 1), C(4, 0).
Let us assume O(x, y) be the Circum - centre of the circle.
We know that distance from circum - centre to any vertex is equal.
So, OA = OB = OC
We know that distance between two points (x1, y1) and (x2, y2) is
Now,
⇒ OA = OB
⇒ OA2 = OB2
⇒ (x - (- 2))2 + (y - 3)2 = (x - 2)2 + (y - (- 1))2
⇒ (x + 2)2 + (y - 3)2 = (x - 2)2 + (y + 1)2
⇒ x2 + 4x + 4 + y2 - 6y + 9 = x2 - 4x + 4 + y2 + 2y + 1
⇒ 8x - 8y = - 8
⇒ x - y = - 1 ..... (1)
Now,
⇒ OB = OC
⇒ OB2 = OC2
⇒ (x - 2)2 + (y - (- 1))2 = (x - 4)2 + (y - 0)2
⇒ (x - 2)2 + (y + 1)2 = (x - 4)2 + (y)2
⇒ x2 - 4x + 4 + y2 + 2y + 1 = x2 - 8x + 16 + y2
⇒ 4x + 2y = 11 .... - (2)
On solving (1) and (2), we get
⇒ and
∴ is the centre of the circle.
We know radius is the distance between the centre and any point on the circle.
Let ‘r’ be the circum - radius of the circle.
∴ The radius of the circle is .
Find the centre of a circle passing through the points (6, - 6), (3, - 7) and (3, 3).
Given that circle passes through the points A(6, - 6), B(3, - 7), C(3, 3).
Let us assume O(x, y) be the centre of the circle.
We know that distance from the centre to any point on h circle is equal.
So, OA = OB = OC
We know that distance between two points (x1, y1) and (x2, y2) is
Now,
⇒ OA = OB
⇒ OA2 = OB2
⇒ (x - 6)2 + (y - (- 6))2 = (x - 3)2 + (y - (- 7))2
⇒ (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2
⇒ x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49
⇒ 6x + 2y = 14
⇒ 3x + y = 7 ..... (1)
Now,
⇒ OB = OC
⇒ OB2 = OC2
⇒ (x - 3)2 + (y - (- 7))2 = (x - 3)2 + (y - 3)2
⇒ (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2
⇒ x2 - 6x + 9 + y2 + 14y + 49 = x2 - 6x + 9 + y2 - 6y + 9
⇒ 20y = - 40
⇒ y = - 2 .... - (2)
Substituting (2) in (1), we get
⇒ x = 3
∴ (3, - 2) is the centre of the circle.
We know radius is the distance between the centre and any point on the circle.
Let ‘r’ be the radius of the circle.
⇒ r = √(9 + 16)
⇒ r = √25
⇒ r = 5
∴ The radius of the circle is 5.
If the line segment joining the points A(a, b) and B(c, d) subtends a right angle at the origin, show that ac + bd = 0.
Given that the line segment joining the points A(a, b) and B(c, d) subtends a right angle at the origin O(0, 0)
So, AOB is a right angled triangle with right angle at O.
We got OA2 + OB2 = AB2 [By Pythagoras Theorem]
We know that the distance between the points (x1, y1) and (x2, y2) is .
⇒ OA2 + OB2 = AB2
⇒ (0 - a)2 + (0 - b)2 + (0 - c)2 + (0 - d)2 = (a - c)2 + (b - d)2
⇒ a2 + b2 + c2 + d2 = a2 + c2 - 2ac + b2 + d2 - 2bd
⇒ 2ac + 2bd = 0
⇒ ac + bd = 0
The centre of the circle is (2x - 1, 3x + 1) and radius is 10 units. Find the value of x if the circle passes through the point (- 3, - 1).
Given that the circle has centre O(2x - 1, 3x + 1) and passes through the point A(- 3, - 1) and has a radius(r) of 10 units.
We know that the radius of the circle is the distance between the centre and any point on the circle.
So, we have r = OA
⇒ OA = 10
⇒ OA2 = 100
⇒ (2x - 1 - (- 3))2 + (3x + 1 - (- 1))2 = 100
⇒ (2x + 2)2 + (3x + 2)2 = 100
⇒ 4x2 + 8x + 4 + 9x2 + 12x + 4 = 100
⇒ 13x2 + 20x - 92 = 0
⇒ 13x2 - 26x + 46x - 92 = 0
⇒ 13x(x - 2) + 46(x - 2) = 0
⇒ (13x + 46)(x - 2) = 0
⇒ 13x + 46 = 0 (or) x - 2 = 0
⇒ 13x = - 46 (or) x = 2
∴ The values of the x are or 2.
Prove that the points (4, 3), (6, 4), (5, 6) and (3, 5) are the vertices of a square.
Given points are A(4, 3), B(6, 4), C(5, 6) and D(3, 5).
We need to prove that these are the vertices of a square.
We know that in the lengths of all sides are equal and the lengths of the diagonals are equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is .
Now,
⇒ AB = √5
⇒ BC = √5
⇒ CD = √5
⇒ DA = √5
We got AB = BC = CD = DA, this may be square (or) rhombus.
Now we find the lengths of the diagonals.
⇒ AC = √10
⇒ BD = √10
We got AC = BD.
∴ The points form a square.
Prove that the points (4, 3), (6, 4), (5, 6) and (- 4, 4) are the vertices of a square.
Given points are A(4, 3), B(6, 4), C(5, 6) and D(- 4, 4).
We need to prove that these are the vertices of a square.
We know that in the lengths of all sides are equal and the lengths of the diagonals are equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is .
Now,
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We got AB = BC≠CD≠DA,
∴ The points doesn’t form a square.
Prove that the points (3, 2), (6, 3), (7, 6), (4, 5) are the vertices of a parallelogram. Is it a rectangle?
Given points are A(3, 2), B(6, 3), C(7, 6) and D(4, 5).
We need to prove that these are the vertices of a parallelogram.
We know that in the lengths of opposite sides are equal in a parallelogram.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is .
Now,
⇒ AB = √10
⇒ BC = √10
⇒ CD = √10
⇒ DA = √10
We got AB = CD and BC = DA, these are the vertices of a parallelogram.
Now we find the lengths of the diagonals.
⇒ AC = √32
⇒ BD = √8
We got AC≠BD.
∴ The points doesn’t form a rectangle.
Prove that the points (6, 8), (3, 7), (- 2, - 2), (1, - 1) are the vertices of a parallelogram.
Given points are A(6, 8), B(3, 7), C(- 2, - 2) and D(1, - 1).
We need to prove that these are the vertices of a parallelogram.
We know that in the lengths of opposite sides are equal in a parallelogram and the lengths of diagonals are not equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is .
Now,
⇒
⇒
⇒
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We got AB = CD and BC = DA, these are the vertices of a parallelogram or rectangle.
Now we find the lengths of the diagonals.
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
We got AC≠BD.
∴ The points form a parallelogram.
Prove that the points (4, 8), (0, 2), (3, 0) and (7, 6) are the vertices of a rectangle.
Given points are A(4, 8), B(0, 2), C(3, 0) and D(7, 6).
We need to prove that these are the vertices of a rectangle.
We know that in the lengths of opposite sides and lengths of diagonals are equal in a rectangle.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is .
Now,
⇒ AB = √52
⇒ BC = √13
⇒ CD = √52
⇒ DA = √13
We got AB = CD and BC = DA, these are the vertices of a parallelogram or a rectangle.
Now we find the lengths of the diagonals.
We got AC = BD.
∴ The points form a rectangle.
Show that the points A(1, 0), B(5, 3), C(2, 7) and D(- 2, 4) are the vertices of a rhombus.
Given points are A(1, 0), B(5, 3), C(2, 7) and D(- 2, 4).
We need to prove that these are the vertices of a rhombus.
We know that in the lengths of sides are equal in a rhombus and the length of diagonals are not equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is .
Now,
⇒
⇒
⇒
⇒
⇒ AB = 5
⇒
⇒
⇒
⇒
⇒ BC = 5
⇒
⇒
⇒
⇒
⇒ CD = 5
⇒
⇒
⇒
⇒
⇒ DA = 5
We got AB = BC = CD = DA, these are the vertices of a square or a rhombus.
Now we find the lengths of the diagonals.
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⇒
⇒
⇒
⇒
⇒
⇒
⇒
We got AC = BD.
∴ The points form a square not rhombus.
Name the type or quadrilateral formed, if any, by the following points and give reasons for your answer:
(4, 5), (7, 6), (4, 3), (1, 2)
Given points are A(4, 5), B(7, 6), C(4, 3) and D(1, 2).
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is .
Now,
⇒
⇒
⇒
⇒
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⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
We got AB = CD and BC = DA, this may be a parallelogram or a rectangle.
Now we find the lengths of the diagonals.
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
We got AC≠BD.
∴ The points form a parallelogram.
Name the type or quadrilateral formed, if any, by the following points and give reasons for your answer:
(- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)
Given points are A(- 1, - 2), B(1, 0), C(- 1, 2) and D(- 3, 0).
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is .
Now,
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⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
We got AB = BC = CD = DA, this may be square (or) rhombus.
Now we find the lengths of the diagonals.
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
We got AC = BD.
∴ The points form a square.
Name the type or quadrilateral formed, if any, by the following points and give reasons for your answer:
(- 3, 5), (3, 1), (0, 3), (- 1, - 4)
Given points are A(- 3, 5), B(3, 1), C(0, 3) and D(- 1, - 4).
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is .
Now,
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⇒
⇒
⇒
⇒
⇒
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⇒
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We got AB≠BC≠CD≠DA, this may be a quadrilateral which is not of standard shape.
Now we find the lengths of the diagonals.
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒ AC + BC = AB
We got points ABC are collinear.
∴ The points doesn’t form a quadrilateral.
Two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of other two vertices.
Given that A(- 1, 2) and C(3, 2) are the opposite vertices of a square.
Let us assume the other two vertices be B(x1, y1) and D(x2, y2) and the midpoint be M
We know that midpoint of AC = Midpoint of BD = M
⇒ M = (1, 2)
⇒ x1 + x2 = 2 ..... (1)
⇒ y1 + y2 = 4 ..... (2)
We know that lengths of the sides of the square are equal.
AB = BC = CD = DA.
We know that distance between two points (x1, y1) and (x2, y2) is
⇒ AB = BC
⇒ AB2 = BC2
⇒ (x1 - (- 1))2 + (y1 - 2)2 = (x1 - 3)2 + (y1 - 2)2
⇒ x12 + 1 + 2x1 + y12 + 4 - 4y1 = x12 - 6x1 + 9 + y12 + 4 - 4y1
⇒ 8x1 = 8
⇒ x1 = 1 ..... (3)
From (1)
⇒ x2 = 2 - 1 = 1 ..... (4)
We know that points ABC form right angled isosceles triangle.
We have AB2 + BC2 = AC2
⇒ 2AB2 = (- 1 - 3)2 + (2 - 2)2
⇒ 2((1 - (- 1))2 + (y1 - 2)2) = (- 4)2 + (0)2
⇒ 2(22 + (y1 - 2)2) = 8
⇒ 4 + (y1 - 2)2 = 8
⇒ (y1 - 2)2 = 4
⇒ y1 - 2 = ±2
⇒ y1 = 2 - 2 (or) y1 = 2 + 2
⇒ y1 = 0 (or) y1 = 4
From (2)
⇒ y2 = 4 - 0
⇒ y2 = 4
⇒ y2 = 4 - 4
⇒ y2 = 0
It is clear that the other two points are (1, 0) and (1, 4).
∴ The other two points are (1, 0) and (1, 4).
If ABCD be a rectangle and P be any point in a plane of the rectangle, then prove that PA2 + PC2 = PB2 + PD2.
[Hint: Take A as the origin and AB and AD as x and y - axis respectively. Let AB = a, AD = b]
Let us assume A as the origin (0, 0) and AB and AD as x and y axis with length a and b units.
Then we get points B to be (a, 0), D to be (0, b) and C to be (a, b).
Let us assume P(x, y) be any point in a plane of the rectangle.
We need to prove PA2 + PC2 = PB2 + PD2.
We know that distance between two points (x1, y1) and (x2, y2) is .
Let us assume L.H.S,
⇒ PA2 + PC2 = ((x - 0)2 + (y - 0)2) + ((x - a)2 + (y - b)2)
⇒ PA2 + PC2 = x2 + y2 + x2 - 2ax + a2 + y2 - 2by + b2
⇒ PA2 + PC2 = (x2 - 2ax + a2 + y2) + (x2 + y2 - 2by + b2)
⇒ PA2 + PC2 = ((x - a)2 + (y - 0)2) + ((x - 0)2 + (y - b)2)
⇒ PA2 + PC2 = PB2 + PD2
⇒ L.H.S = R.H.S
∴ Thus proved.
Prove, using co - ordinates that diagonals of a rectangle are equal.
Let us assume ABCD be a rectangle with A as the origin and AB and AD as x and y - axes having lengths a and b units.
We get the vertices of the rectangle as follows.
⇒ A = (0, 0)
⇒ B = (a, 0)
⇒ C = (a, b)
⇒ D = (0, b)
We need to prove the lengths of the diagonals are equal.
i.e., AC = BD
We know that distance between two points (x1, y1) and (x2, y2) is .
Let us find the individual lengths of diagonals,
⇒
⇒ ..... (1)
⇒
⇒ ..... (2)
From (1) and (2), we can clearly say that AC = BD.
∴ The diagonals of a rectangle are equal.
Prove, using coordinates that the sum of squares of the diagonals of a rectangle is equal to the sum of squares of its sides.
Let us assume ABCD be a rectangle with A as the origin and AB and AD as x and y - axes having lengths a and b units.
We get the vertices of the rectangle as follows.
⇒ A = (0, 0)
⇒ B = (a, 0)
⇒ C = (a, b)
⇒ D = (0, b)
We need to prove that the sum of squares of the diagonals of a rectangle is equal to the sum of squares of its sides.
i.e., AC2 + BD2 = AB2 + BC2 + CD2 + DA2
We know that distance between two points (x1, y1) and (x2, y2) is .
Assume L.H.S
⇒ AC2 + BD2 = ((0 - a)2 + (0 - b)2) + ((a - 0)2 + (0 - b)2)
⇒ AC2 + BD2 = a2 + b2 + a2 + b2
⇒ AC2 + BD2 = 2(a2 + b2) ..... - (1)
Assume R.H.S
⇒ AB2 + BC2 + CD2 + DA2 = ((0 - a)2 + (0 - 0)2) + ((a - a)2 + (0 - b)2) + ((a - 0)2 + (b - b)2) + ((0 - 0)2 + (b - 0)2)
⇒ AB2 + BC2 + CD2 + DA2 = a2 + 0 + 0 + b2 + a2 + 0 + 0 + b2
⇒ AB2 + BC2 + CD2 + DA2 = 2(a2 + b2) .... (2)
From (1) and (2), we can clearly say that,
⇒ AC2 + BD2 = AB2 + BC2 + CD2 + DA2
∴ Thus proved.
Find the coordinates of the point which divides the line segment joining (2,4) and (6,8) in the ratio 1:3 internally and externally.
Let P(x,y) be the point which divides the line segment internally.
Using the section formula for the internal division, i.e.
…(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (2, 4) and (x2, y2) = (6, 8)
Putting the above values in the above formula, we get
⇒ x = 3, y = 5
Hence, (3,5) is the point which divides the line segment internally.
Now, Let Q(x,y) be the point which divides the line segment externally.
Using the section formula for the external division, i.e.
…(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (2, 4) and (x2, y2) = (6, 8)
Putting the above values in the above formula, we get
⇒ x = 0, y = 2
Hence, (0,2) is the point which divides the line segment externally.
Find the coordinates of the point which divides the join of (-1,7) and (4,-3) internally in the ratio 2:3.
Let P(x,y) be the point which divides the line segment internally.
Using the section formula for the internal division, i.e.
…(i)
Here, m1 = 2, m2 = 3
(x1, y1) = (-1, 7) and (x2, y2) = (4, -3)
Putting the above values in the above formula, we get
⇒ x = 1, y = 3
Hence, (1,3) is the point which divides the line segment internally.
Find the coordinates of the point which divides the line segment joining the points (4,-3) and (8,5) in the ratio 3:1 internally.
Let P(x,y) be the point which divides the line segment internally.
Using the section formula for the internal division, i.e.
…(i)
Here, m1 = 3, m2 = 1
(x1, y1) = (4, -3) and (x2, y2) = (8, 5)
Putting the above values in the above formula, we get
⇒ x = 7, y = 3
Hence, (7,3) is the point which divides the line segment internally.
Find the coordinates of the points which trisect the line segment joining the points (2,3) and (6,5).
Let P and Q be the points of trisection of AB, i.e. AP = PQ = QB
∴ P divides AB internally in the ratio 1: 2.
∴ the coordinates of P, by applying the section formula, are
…(i)
Here, m1 = 1, m2 = 2
(x1, y1) = (2, 3) and (x2, y2) = (6, 5)
Putting the above values in the above formula, we get
Now, Q also divides AB internally in the ratio 2: 1. So, the coordinates of Q are
…(i)
Here, m1 = 2, m2 = 1
(x1, y1) = (2, 3) and (x2, y2) = (6, 5)
Putting the above values in the above formula, we get
Therefore, the coordinates of the points of trisection of the line segment joining A and B are
Find the coordinates of the point of trisection of the line segment joining (1,-2) and (-3,4).
Let P and Q be the points of trisection of AB, i.e. AP = PQ = QB
∴ P divides AB internally in the ratio 1: 2.
∴ the coordinates of P, by applying the section formula, are
…(i)
Here, m1 = 1, m2 = 2
(x1, y1) = (1, -2) and (x2, y2) = (-3, 4)
Putting the above values in the above formula, we get
Now, Q also divides AB internally in the ratio 2: 1. So, the coordinates of Q are
…(i)
Here, m1 = 2, m2 = 1
(x1, y1) = (1, -2) and (x2, y2) = (-3, 4)
Putting the above values in the above formula, we get
Therefore, the coordinates of the points of trisection of the line segment joining A and B are
The coordinates of A and B are (1,2) and (2,3) respectively, If P lies on AB, find the coordinates of P such that
Given:
⇒ m1 = 4 and m2 = 3
and (x1, y1) = (1, 2) ; (x2, y2) = (2, 3)
Using the section formula for the internal division, i.e.
…(i)
Hence, the coordinates of P are
If A (4,-8), B (3,6) and C(5,-4) are the vertices of a ΔABC, D is the mid-point of BC and P is a point on AD joined such that , find the coordinates of P.
Given: D is the midpoint of BC. So, BD = DC
Then the coordinates of D are
⇒ x = 4 and y = 1
So, coordinates of D are (4, 1)
Now, we have to find the coordinates of P.
Given:
⇒ m1 = 2 and m2 = 1
and (x1, y1) = (4, -8) ; (x2, y2) = (4, 1)
Using the section formula for the internal division, i.e.
…(i)
⇒ x = 4, y = -2
Hence, the coordinates of P are P(x,y) = P(4, -2)
If p divides the join of A (-2,-2) and B (2,-4) such that , find the coordinates of P.
Given:
⇒ 7AP = 3AP + 3PB
⇒ 7AP – 3AP = 3PB
⇒ 4AP = 3PB
Hence, the point P divides AB in the ratio of 3:4
⇒ m1 = 3 and m2 = 4
and (x1, y1) = (-2, -2) ; (x2, y2) = (2, -4)
Using the section formula for the internal division, i.e.
…(i)
Hence, the coordinates of P are
A (1,4) and B (4,8) are two points. P is a point on AB such that AP = AB + BP. If AP = 10 find the coordinates of P.
Given: AP = AB + BP and AP = 10
Firstly, we find the distance between A and B
d(A,B) = √(x2 – x1)2 + (y2 – y1)2
= √(4 – 1)2 + (8 – 4)2
= √(3)2 + (4)2
= √9 + 16
= √25
= 5
So, AB = 5
It is given that AP = AB + BP
⇒ 10 = 5 + BP
⇒ 10 – 5 = BP
⇒ BP = 5
⇒ A, B and P are collinear
and since AB = BP
⇒ B is the midpoint of AP
Let the coordinates of P = (x,y)
⇒ x + 1 = 8 and y + 4 = 16
⇒ x = 7 and y = 12
Hence, the coordinates of P are (7, 12)
The line segment joining A (2,3) and B(-3,5) is extended through each end by a length equal to its original length. Find the coordinates of the new ends.
Let P and Q be the required new ends
Coordinates of P
Let AP = k
∴ AB = AP = k
and PB = AP + AB = k + k = 2k
∴ P divides AB externally in the ratio 1:2
Using the section formula for the external division, i.e.
…(i)
Here, m1 = 1, m2 = 2
(x1, y1) = (2, 3) and (x2, y2) = (-3, 5)
Putting the above values in the above formula, we get
⇒ x = 7, y = 1
∴Coordinates of P are (7, 1)
Coordinates of Q.
Q divides AB externally in the ratio 2:1
Again, Using the section formula for the external division, i.e.
…(i)
Here, m1 = 2, m2 = 1
(x1, y1) = (2, 3) and (x2, y2) = (-3, 5)
Putting the above values in the above formula, we get
∴Coordinates of Q are (-8, 7)
The line segment joining A(6,3) to B(-1,-4) is doubled in length by having half its length added to each end. Find the coordinates of the new ends.
Let P and Q be the required new ends
Coordinates of P
Let AP = k
∴ AB = 2AP = 2k
and PB = AP + AB = k + 2k = 3k
∴ P divides AB externally in the ratio 1:3
Using the section formula for the external division, i.e.
…(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (6, 3) and (x2, y2) = (-1, -4)
Putting the above values in the above formula, we get
∴Coordinates of P are
Coordinates of Q.
Q divides AB externally in the ratio 3:1
Again, Using the section formula for the external division, i.e.
…(i)
Here, m1 = 3, m2 = 1
(x1, y1) = (6, 3) and (x2, y2) = (-1, -4)
Putting the above values in the above formula, we get
∴Coordinates of Q are
The coordinates of two points A and B are (-1,4) and (5,1) respectively. Find the coordinates of the point P which lies on extended line AB such that it is three times as far from B as from A.
Now, Let P(x,y) be the point which lies on extended line AB
Using the section formula for the external division, i.e.
…(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (-1, 4) and (x2, y2) = (5, 1)
Putting the above values in the above formula, we get
Find the distances of that point from the origin which divides the line segment joining the points (5,-4) and (3,-2) in the ration 4:3.
Let the coordinates of the point be (x,y)
Let A = (5, -4) and B = (3, -2)
Here, the point divides the line segment in the ratio 4:3
So, m1 = 4 and m2 = 3
Using section formula,
…(i)
Hence, the coordinates of P are
Now, the distance from the origin (0,0) is
The coordinates of the middle points of the sides of a triangle are (1,1), (2,3) and (4,1), find the coordinates of its vertices.
Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(1, 1), Q(2, 3) and R(4, 1) are the midpoints of AB, BC, and CA. Then,
…(i)
…(ii)
…(iii)
…(iv)
…(v)
…(vi)
Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 2 + 4 + 8
⇒ 2(x1 + x2 + x3) = 14
⇒ x1 + x2 + x3 = 7 …(vii)
From (i) and (vii), we get
x3 = 7 – 2 = 5
From (iii) and (vii), we get
x1 = 7 – 4 = 3
From (v) and (vii), we get
x2 = 7 – 8 = -1
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 2 + 6 + 2
⇒ 2(y1 + y2 + y3) = 10
⇒ y1 + y2 + y3 = 5 …(viii)
From (ii) and (viii), we get
y3 = 5 – 2 = 3
From (iv) and (vii), we get
y1 = 5 – 6 = -1
From (vi) and (vii), we get
y2 = 5 – 2 = 3
Hence, the vertices of ΔABC are A(3, -1), B(-1, 3) and C(5, 3)
If the points (10,5),(8,4) and (6,6) are the mid-points of the sides of a triangle, find its vertices.
Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(10, 5), Q(8, 4) and R(6, 6) are the midpoints of AB, BC, and CA. Then,
…(i)
…(ii)
…(iii)
…(iv)
…(v)
…(vi)
Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 20 + 16 + 12
⇒ 2(x1 + x2 + x3) = 48
⇒ x1 + x2 + x3 = 24 …(vii)
From (i) and (vii), we get
x3 = 24 – 20 = 4
From (iii) and (vii), we get
x1 = 24 – 16 = 8
From (v) and (vii), we get
x2 = 24 – 12 = 12
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 10 + 8 + 12
⇒ 2(y1 + y2 + y3) = 30
⇒ y1 + y2 + y3 = 15 …(viii)
From (ii) and (viii), we get
y3 = 15 – 10 = 5
From (iv) and (vii), we get
y1 = 15 – 8 = 7
From (vi) and (vii), we get
y2 = 15 – 12 = 3
Hence, the vertices of ΔABC are A(8, 7), B(12, 3) and C(4, 5)
The mid-points of the sides of a triangle are (3,4),(4,6) and (5,7). Find the coordinates of the vertices of the triangle.
Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(3, 4), Q(4, 6) and R(5, 7) are the midpoints of AB, BC, and CA. Then,
…(i)
…(ii)
…(iii)
…(iv)
…(v)
…(vi)
Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 6 + 8 + 10
⇒ 2(x1 + x2 + x3) = 24
⇒ x1 + x2 + x3 =12 …(vii)
From (i) and (vii), we get
x3 = 12 – 6 = 6
From (iii) and (vii), we get
x1 = 12 – 8 = 4
From (v) and (vii), we get
x2 = 12 – 10 = 2
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 8 + 12 + 14
⇒ 2(y1 + y2 + y3) = 34
⇒ y1 + y2 + y3 = 17 …(viii)
From (ii) and (viii), we get
y3 = 17 – 8 = 9
From (iv) and (vii), we get
y1 = 17 – 12 = 5
From (vi) and (vii), we get
y2 = 17 – 14 = 3
Hence, the vertices of ΔABC are A(4, 5), B(2, 3) and C(6, 9)
A(1,-2) and B(2,5) are two points. The lines OA, OB are produced to C and D respectively such that OC = 2OA and OD = 2OB. Find CD.
Given:
A(1, -2) and B(2, 5) are two points.
OC = 2OA …(i)
and OD = 2OB …(ii)
Adding (i) and (ii), we get
OC + OD = 2OA + 2OB
⇒ CD = 2[OA + OB]
⇒ CD = 2[AB] …(iii)
Now, we find the distance between A and B
d(A,B) = √(x2 – x1)2 + (y2 – y1)2
= √(2 – 1)2 + {5 – (-2)}2
= √(1)2 + (5 + 2)2
= √1 + 49
= √50
= 5√2
Putting the value in eq. (iii), we get
CD = 2 × 5√2
= 10√2
Find the length of the medians of the triangle whose vertices are (-1,3),(1,-1) and (5,1).
Let the given points of a triangle be A(-1, 3), B(1, -1) and C(5,1)
Let D, E and F are the midpoints of the sides BC, CA and AB respectively.
The coordinates of D are:
D = (3, 0)
The coordinates of E are:
E = (2, 2)
The coordinates of F are:
F = (0, 1)
Now, we have to find the lengths of the medians.
d(A,D) = √(x2 – x1)2 + (y2 – y1)2
= √{3 – (-1)2} + {0 – 3}2
= √(3 + 1)2 + (-3)2
= √16 + 9
= √25
= 5 units
d(B,E) = √(x2 – x1)2 + (y2 – y1)2
= √(2 – 1)2 + {2 – (-1)}2
= √(1)2 + (2 + 1)2
= √1 + 9
= √10 units
d(C,F) = √(x2 – x1)2 + (y2 – y1)2
= √(5 – 0)2 + {1 – 1}2
= √(5)2 + (0)2
= √25
= 5 units
Hence, the length of the medians AD, BE and CF are 5, √10, 5 units respectively.
If A(1,5), B (-2,1) and C(4,1) be the vertices of ΔABC and the internal bisector of ∠A meets BC and D, find AD.
Given: A(1, 5), B(-2, 1) and C(4,1) are the vertices of ΔABC
Using angle bisector theorem, which states that:
The ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC:{\displaystyle {\frac {|BD|}{|DC|}}={\frac {|AB|}{|AC|}},}
⇒ BD = DC
⇒ D is the midpoint of BC
So, the coordinates of D are:
D = (1, 1)
Now, AD = √(x2 – x1)2 + (y2 – y1)2
= √(1 – 1)2 + {5 – 1}2
= √(0)2 + (4)2
= √16
= 4 units
Hence, AD = 4 units
If the middle point of the line segment joining (3,4) and (k,7) is (x,y) and 2x+2y+1=0, find the value of k.
Let P be the midpoint of the line segment joining (3, 4) and (k, 7)
So, the coordinates of P are:
Again,
2x + 2y + 1 = 0
⇒ 3 + k + 11 + 1 = 0
⇒ k + 15 = 0
⇒ k = -15
one end of a diameter of a circle is at (2,3) and the center is (-2,5), find the coordinates of the other end of the diameter.
Let the coordinates of the other end be (x,y).
Since (-2, 5) is the midpoint of the line joining (2,3) and (x,y)
⇒ x + 2 = -4 and y + 3 = 10
⇒ x = -4 – 2 and y = 10 – 3
⇒ x = -6 and y = 7
Hence, the coordinates of the other end are (-6, 7)
Find the coordinates of a point A, where AB is the diameter of a circle whose center is (2,-3), and B is (1,4)
Let the coordinates of the A be (x,y).
Since 2, -3) is the midpoint of the line joining (1, 4) and (x,y)
⇒ x + 1 = 4 and y + 4 = -6
⇒ x = 4 – 1 and y = -6 – 4
⇒ x = 3 and y = -10
Hence, the coordinates of A are (3, -10)
If the point C (-1,2) divides internally the line segment joining A (2,5) and B in the ratio 3:4. Find the coordinates of B.
Let the coordinates of B are (x, y)
It is given that the line segment divide in the ratio 3:4
So, m1 = 3 and m2 = 4
and (x’, y’) =(-1, 2); (x1, y1) = (2,5); (x2, y2) = (x, y)
Using section formula for the internal division, we get
…(i)
⇒ 3x + 8 = -7 and 3y + 20 = 14
⇒ 3x = -7 – 8 and 3y = 14 – 20
⇒ 3x = -15 and 3y = -6
⇒ x = -5 and y = -2
Hence, the coordinates of B are (-5, -2)
Find the ratio in which (-8,3) divides the line segment joining the points (2,-2) and (-4,1).
Let C(-8, 3) divides the line segment AB in the ratio m:n
Here, (x, y) = (-8, 3); (x1, y1) = (2, -2) and (x2, y2) = (-4,1)
So,
⇒ -8m -8n = -4m + 2n and 3m + 3n = m - 2n
⇒ -8m + 4m - 8n - 2n = 0 and 3m – m + 3n + 2n = 0
⇒ -4m – 10n = 0 and 2m + 5n = 0
⇒ -2m – 5n = 0 and 2m + 5n = 0
⇒ 2m + 5n = 0 and 2m + 5n = 0
⇒ 2m = -5n
Hence, the ratio is 5:2 and this negative sign shows that the division is external.
In what ratio does the point (-4,6) divide the line segment joining the point A(-6,10) and B (3,-8)?
Let C(-4, 6) divides the line segment AB in the ratio m:n
Here, (x, y) = (-4, 6); (x1, y1) = (-6, 10) and (x2, y2) = (3, -8)
So,
⇒ -4m -4n = 3m - 6n and 6m + 6n = -8m + 10n
⇒ -4m – 3m – 4n + 6n = 0 and 6m + 8m + 6n – 10n = 0
⇒ -7m + 2n = 0 and 14m - 4n = 0
⇒ -7m = -2n and 14m = 4n
⇒ 7m = 2n and 7m = 2n
Hence, the ratio is 2:7 and the division is internal.
Find the ratio in which the line segment joining (-3,10) and (6,-8) is divided by (-1,6)
Let C(-1, 6) divides the line segment AB in the ratio m:n
Here, (x, y) = (-1, 6); (x1, y1) = (-3, 10) and (x2, y2) = (6, -8)
So,
⇒ -m – n = 6m - 3n and 6m + 6n = -8m + 10n
⇒ -m – 6m – n + 3n = 0 and 6m + 8m + 6n – 10n = 0
⇒ -7m + 2n = 0 and 14m - 4n = 0
⇒ 2n = 7m and 4n = 14m
⇒ 7m = 2n
Hence, the ratio is 2:7 and the division is internal.
Find the ratio in which the line segment joining (-3,-4) and (3,5) is divided by (x,2). Also, find x.
Let C(x, 2) divides the line segment AB in the ratio m:n
Here, (x, y) = (x, 2); (x1, y1) = (-3, -4) and (x2, y2) = (3, 5)
So,
⇒ 2m + 2n = 5m – 4n
⇒ 2m – 5m = -4n – 2n
⇒ -3m = -6n
⇒ m = 2n
Now, the ratio is 2:1
Now,
⇒ 3x = 6 – 3
⇒ 3x = 3
⇒ x = 1
Hence, the ratio is 2:1 and the division is internal and the value of x = 1
In what ratio does the x-axis divide the line segment joining the points (2,-3) and (5,6).
Let the line segment A(2, -3) and B(5, 6) is divided at point P(x,0) by x-axis in ratio m:n
Here, (x, y) = (x, 0); (x1, y1) = (2, -3) and (x2, y2) = (5, 6)
So,
⇒ 0 = 6m – 3n
⇒ -6m = -3n
Hence, the ratio is 1:2 and the division is internal.
Find the ratio in which the line segment joining A(1,-5) and B(-4,5) is divided by the x-axis. Also, find the coordinates of the point of division.
Let the line segment A(1, -5) and B(-4, 5) is divided at point P(x,0) by x-axis in ratio m:n
Here, (x, y) = (x, 0); (x1, y1) = (1, -5) and (x2, y2) = (-4, 5)
So,
⇒ 0 = 5m – 5n
⇒ 5m = 5n
Hence, the ratio is 1:1 and the division is internal.
Now,
Hence, the coordinates of the point of division is
Find the ratio in which the y-axis divides the line segment joining points (5,-6) and (-1,-4). Also, find the point of intersection.
Let the line segment A(5, -6) and B(-1, -4) is divided at point P(0, y) by y-axis in ratio m:n
Here, (x, y) = (0, y); (x1, y1) = (5, -6) and (x2, y2) = (-1, -4)
So,
⇒ 0 = -m + 5n
⇒ m = 5n
Hence, the ratio is 5:1 and the division is internal.
Now,
Hence, the coordinates of the point of division is
Find the centroid of the triangle whose vertices are (2,4), (6,4), (2,0).
Here, x1 = 2, x2 = 6, x3 = 2
and y1 = 4, y2 = 4, y3 = 0
Let the coordinates of the centroid be(x,y)
So,
Hence, the centroid of a triangle is
The vertices of a triangle are at (2,2), (0,6) and (8,10). Find the coordinates of the trisection point of each median which is nearer the opposite side.
Let (2, 2), (0, 6) and (8, 10) be the vertices A, B and C of the triangle respectively. Let AD, BE, CF be the medians
The coordinates of D are:
D = (4, 8)
The coordinates of E are:
E = (5, 6)
The coordinates of F are:
F = (1, 4)
Let P be the trisection point of the median AD which is nearer to the opposite side BC
∴ P divides DA in the ratio 1:2 internally
Let Q be the trisection point of the median BE which is nearer to the opposite side CA
∴ Q divides EB in the ratio 1:2 internally
Let R be the trisection point of the median CF which is nearer to the opposite side AB
∴ R divides FC in the ratio 1:2 internally
Therefore, Coordinates of required trisection points are
Two vertices of a triangle are (1,4) and (5,2). If its centroid is (0,-3), find the third vertex.
Let the third vertex of a triangle be(x,y)
Here, x1 = 1, x2 = 5, x3 = x
and y1 = 4, y2 = 2, y3 = y
and the coordinates of the centroid is (0, -3)
We know that
⇒ 6 + x = 0 and 6 + y = -9
⇒ x = -6 and y = -15
Hence, the third vertex of a triangle is (-6, -15)
The coordinates of the centroid of a triangle are (√3,2), and two of its vertices are (2√3,-1) and (2√3,5). Find the third vertex of the triangle.
Let the third vertex of a triangle be (x, y)
Here, x1 = 2√3, x2 = 2√3, x3 = x
and y1 = -1, y2 = 5, y3 = y
and the coordinates of the centroid is (√3, 2)
We know that
⇒ 4√3 + x = 3√3 and 4 + y = 6
⇒ x = -√3 and y = 2
Hence, the third vertex of a triangle is (-√3, 2)
Find the centroid of the triangle ABC whose vertices are A (9,2), B(1,10) and C(-7,-6). Find the coordinates of the middle points of its sides and hence find the centroid of the triangle formed by joining these middle points. Do the two triangles have the same centroid?
The vertices of a triangle are A (9,2), B(1,10) and C(-7,-6)
Here, x1 = 9, x2 = 1, x3 = -7
and y1 = 2, y2 = 10, y3 = -6
Let the coordinates of the centroid be(x,y)
So,
= (1,2)
Hence, the centroid of a triangle is (1, 2)
Now,
Let D, E and F are the midpoints of the sides BC, CA and AB respectively.
The coordinates of D are:
D = (-3, 2)
The coordinates of E are:
E = (1, -2)
The coordinates of F are:
F = (5, 6)
Now, we find the centroid of a triangle formed by joining these middle points D, E, and F as shown in figure
Let P be the trisection point of the median AD which is nearer to the opposite side BC
∴ P divides DA in the ratio 1:2 internally
= (1, 2)
Let Q be the trisection point of the median BE which is nearer to the opposite side CA
∴ Q divides EB in the ratio 1:2 internally
= (1, 2)
Let R be the trisection point of the median CF which is nearer to the opposite side AB
∴ R divides FC in the ratio 1:2 internally
= (1, 2)
Yes, the triangle has the same centroid, i.e. (1,2)
If (1,2), (0,-1) and (2,-1) are the middle points of the sides of the triangle, find the coordinates of its centroid.
Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(1, 2), Q(0, -1) and R(2, -1) are the midpoints of AB, BC and CA. Then,
…(i)
…(ii)
…(iii)
…(iv)
…(v)
…(vi)
Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 2 + 0 + 4
⇒ 2(x1 + x2 + x3) = 6
⇒ x1 + x2 + x3 = 3 …(vii)
From (i) and (vii), we get
x3 = 3 – 2 = 1
From (iii) and (vii), we get
x1 = 3 – 0 = 3
From (v) and (vii), we get
x2 = 3 – 4 = -1
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 4 + (-2) + (-2)
⇒ 2(y1 + y2 + y3) = 0
⇒ y1 + y2 + y3 = 0 …(viii)
From (ii) and (viii), we get
y3 = 0 – 4 = -4
From (iv) and (vii), we get
y1 = 0 – (-2) = 2
From (vi) and (vii), we get
y2 = 0 – (-2) = 2
Hence, the vertices of ΔABC are A(3, 2), B(-1, 2) and C(1, -4)
Now, we have to find the centroid of a triangle
The vertices of a triangle are A(3, 2), B(-1, 2) and C(1, -4)
Here, x1 = 3, x2 = -1, x3 = 1
and y1 = 2, y2 = 2, y3 = -4
Let the coordinates of the centroid be(x,y)
So,
= (1,0)
Hence, the centroid of a triangle is (1, 0)
Show that A(-3,2), B(-5,-5), C(2,-3) and D(4,4) are the vertices of a rhombus.
Note that to show that a quadrilateral is a rhombus, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD have the same midpoint.
(b) a pair of adjacent edges are equal
(c) the diagonal AC and BD are not equal.
Let A(-3, 2), B(-5,-5), C(2,-3) and D(4,4) are the vertices of a rhombus.
Coordinates of the midpoint of AC are
Coordinates of the midpoint of BD are
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, using Distance Formula
d(A,B)= AB = √(-5 + 3)2 + (-5 – 2)2
⇒ AB = √(-2)2 + (-7)2
⇒ AB = √4 +49
⇒ AB = √53 units
d(B,C)= BC = √(-5 – 2)2 + (-5 + 3)2
⇒ BC = √(-7)2 + (-2)2
⇒ BC = √49 +4
⇒ BC = √53 units
d(C,D) = CD = √(4 – 2)2 + (4 + 3)2
⇒ CD = √(2)2 + (7)2
⇒ CD = √4 +49
⇒ CD = √53 units
d(A,D) = AD =√(4 + 3)2 +(4 – 2)2
⇒ AD = √(7)2 + (2)2
⇒ AD = √49 +4
⇒ AD = √53 units
Therefore, AB = BC = CD = AD = √53 units
Now, check for the diagonals
AC = √(2 + 3)2 + (-3 – 2)2
= √(5)2 + (-5)2
= √25 + 25
= √50
and
BD = √(4 + 5)2 + (4 + 5)2
⇒ BD = √(9)2 + (9)2
⇒ BD = √81 + 81
⇒ BD = √162
⇒ Diagonal AC ≠ Diagonal BD
Hence, ABCD is a rhombus.
Show that the point (3,2),(0,5),(-3,2) and (0,-1) are the vertices of a square.
Note that to show that a quadrilateral is a square, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD bisect each other
(b) a pair of adjacent edges are equal
(c) the diagonal AC and BD are equal.
Let the vertices of a quadrilateral are A(3, 2), B(0,5), C(-3, 2) and D(0, -1).
Coordinates of the midpoint of AC are
Coordinates of the midpoint of BD are
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, Using Distance Formula, we get
AB = √(x2 – x1)2 + (y2 – y1)2
= √[(0 – 3)2 + (5 - 2)2]
= √(-3)2 + (3)2
= √(9 + 9)
= √18 units
BC = √[(-3 – 0)2 + (2 - 5)2]
= √(-3)2 + (-3)2
= √(9 + 9)
= √18 units
Therefore, AB = BC = √18 units
Now, check for the diagonals
AC = √(-3 – 3)2 + (2 – 2)2
= √(-6)2 + (0)2
= √36
= 6 units
and
BD = √(0 - 0)2 + (-1 – 5)2
⇒ BD = √(0)2 + (-6)2
⇒ BD = √36
⇒ BD = 6 units
∴ AC = BD
Hence, ABCD is a square.
Prove that the points (-2,-1), (1,0),(4,3) and (1,2) are the vertices of a parallelogram.
Note that to show that a quadrilateral is a parallelogram, it is sufficient to show that the diagonals of the quadrilateral bisect each other.
Let A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram.
Let M be the midpoint of AC, then the coordinates of M are given by
Let N be the midpoint of BD, then the coordinates of N are given by
Thus, AC and BD have the same midpoint.
In other words, AC and BD bisect each other.
Hence, ABCD is a parallelogram.
Show that the points A(1,0), B(5,3), C(2,7) and D(-2,4) are the vertices of a rhombus.
Note that to show that a quadrilateral is a rhombus, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD have the same midpoint.
(b) a pair of adjacent edges are equal
Let A(1, 0), B(5, 3), C(2, 7) and D(-2, 4) are the vertices of a rhombus.
Coordinates of the midpoint of AC are
Coordinates of the midpoint of BD are
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, using Distance Formula
d(A,B)= AB = √(5 – 1)2 + (3 – 0)2
⇒ AB = √(4)2 + (3)2
⇒ AB = √16 + 9
⇒ AB = √25 = 5 units
d(B,C)= BC = √(2 – 5)2 + (7 – 3)2
⇒ BC = √(-3)2 + (4)2
⇒ BC = √9 + 16
⇒ BC = √25 = 5 units
Therefore, adjacent sides are equal.
Hence, ABCD is a rhombus.
Prove that the point (4,8), (0,2), (3,0) and (7,6) are the vertices of a rectangle.
Note that to show that a quadrilateral is a rectangle, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD bisect each other and,
(b) the diagonal AC and BD are equal
Let A(4, 8), B(0, 2), C(3, 0) and D(7, 6) are the vertices of a rectangle.
Coordinates of the midpoint of AC are
Coordinates of the midpoint of BD are
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, check for the diagonals by using the distance formula
AC = √(3 – 4)2 + (0 – 8)2
= √(-1)2 + (-8)2
= √1 + 64
= √65 units
and
BD = √(7 - 0)2 + (6 – 2)2
⇒ BD = √(7)2 + (4)2
⇒ BD = √49 + 16
⇒ BD = √65 units
∴ AC = BD
Hence, ABCD is a rectangle.
Prove that the points (4,3), (6,4), (5,6) and (3,5) are the vertices of a square.
Note that to show that a quadrilateral is a square, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD bisect each other
(b) a pair of adjacent edges are equal
(c) the diagonal AC and BD are equal.
Let the vertices of a quadrilateral are A(4, 3), B(6, 4), C(5, 6) and D(3, 5).
Coordinates of the midpoint of AC are
Coordinates of the midpoint of BD are
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, Using Distance Formula, we get
AB = √(x2 – x1)2 + (y2 – y1)2
= √[(6 – 4)2 + (4 - 3)2]
= √(2)2 + (1)2
= √(4 + 1)
= √5 units
BC = √[(5 – 6)2 + (6 - 4)2]
= √(-1)2 + (2)2
= √(1 + 4)
= √5 units
Therefore, AB = BC = √5 units
Now, check for the diagonals
AC = √(5 – 4)2 + (6 – 3)2
= √(1)2 + (3)2
= √1 + 9
= √10 units
and
BD = √(3 - 6)2 + (5 – 4)2
⇒ BD = √(-3)2 + (1)2
⇒ BD = √9 + 1
⇒ BD = √10 units
∴ AC = BD
Hence, ABCD is a square.
If (6,8), (3,7) and (-2,-2) be the coordinates of the three consecutive vertices of a parallelogram, find coordinates of the fourth vertex.
Let the coordinates of the fourth vertex D be (x, y).
We know that diagonals of a parallelogram bisect each other.
∴ Midpoint of AC = Midpoint of BD …(i)
Coordinates of the midpoint of AC are
Coordinates of the midpoint of BD are
So, according to eq. (i), we have
⇒ 3 + x = 4 and 7 + y = 6
⇒ x = 1 and y = -1
Thus, the coordinates of the vertex D are (1, -1)
Three consecutive vertices of a rhombus are (5,3), (2,7) and (-2,4). Find the fourth vertex.
Let the coordinates of the fourth vertex D be (x, y).
We know that diagonals of a rhombus bisect each other.
∴ Midpoint of AC = Midpoint of BD …(i)
Coordinates of the midpoint of AC are
Coordinates of the midpoint of BD are
So, according to eq. (i), we have
⇒ 2 + x = 3 and 7 + y = 7
⇒ x = 1 and y = 0
Thus, the coordinates of the vertex D are (1, 0)
A quadrilateral has the vertices at the point (-4,2), (2,6), (8,5) and (9,-7). Show that the mid-point of the sides of this quadrilateral are the vertices of a parallelogram.
Let the vertices of quadrilateral be P(-4,2), Q(2,6), R(8,5) and S(9,-7)
Let A, B, C and D are the midpoints of PQ, QR, RS and SP respectively.
Now, since A is the midpoint of P(-4, 2) and Q(2, 6)
∴ Coordinates of A are
Coordinates of B are
Coordinates of C are
and
Coordinates of D are
Now,
we find the distance between A and B
Now, since length of opposite sides of the quadrilateral formed by the midpoints of the given quadrilateral are equal .i.e.
AB = CD and AD = BC
∴ it is a parallelogram
Hence Proved
If the points A(6,1), B(8,2), C(9,4) and D(p,3) are the vertices of a parallelogram taken in order, find the value of p.
Let the points be A(6,1), B(8,2), C(9,4) and D(p,3)
We know that diagonals of parallelogram bisect each other.
∴ Midpoint of AC = Midpoint of BD …(i)
Coordinates of the midpoint of AC are
Coordinates of the midpoint of BD are
So, according to eq. (i), we have
⇒ 8 + p = 15
⇒ p = 15 – 8 = 7
Hence, the value of p is 7
Prove that the line segment joining the middle points of two sides of a triangle is half the third side.
We take O as the origin and OX and OY as the x and y axis respectively.
Let BC = 2a, then B = (-a, 0) and C = (a, 0)
Let A = (b, c), if E and F are the midpoints of sides AC and AB respectively.
Coordinates of midpoint of AC are
Coordinates of the midpoint of AB are
Now, distance between F and E is
d(F,E) = √(x2 – x1)2 + (y2 – y1)2
= a …(i)
and Length of BC = 2a …(ii)
From (i) and (ii), we can say that
Hence Proved
If P,Q,R divide the side BC,CA and AB of ΔABC in the same ratio, prove that the centroid of the triangle ABC and PQR coincide.
Let P, Q, R be the midpoints of sides BC, CA and AB respectively
Construct a ΔPQR by joining these three midpoints of the sides.
This is called the medial triangle
Since, PQ, QR and PR are midsegments of BC, AB and AC respectively
So,
Since the corresponding sides are proportional
∴ ΔPQR ≅ ΔABC
Now, we have to prove that the centroid of the triangle ABC and PQR coincide.
For that we must show that the medians of ΔABC pass through the midpoints of three sides of the medial triangle ΔPQR.
Since PQ is a midsegment of ΔABC,
⇒ PQ || BC, so PQ || BR.
And since QR is a midsegment of AB,
⇒ AB || QR, so QR || PB.
By definition, a quadrilateral PQRB is a parallelogram.
The medians BQ and CP are in fact the diagonals of the parallelogram PQRB.
And we know that the diagonals of a parallelogram bisect each other, so PD = DR.
In other words, D is the midpoint of PR.
In the similar manner, we can show that F and E are midpoints of RQ and PQ respectively.
Hence, the centroid of the triangle ABC and PQR coincide.
Find the area of the triangle whose vertices are
(3, -4), (7, 5), (-1, 10)
Given: (3, -4), (7, 5), (-1, 10)
Let us Assume A(x1, y1) = (3, -4)
Let us Assume B(x2, y2) = (7, 5)
Let us Assume C(x3, y3) = (-1, 10)
Area of triangle
Now,
Area of given triangle
= 46 square units
Find the area of the triangle whose vertices are
(-1.5, 3), (6, -2), (-3, 4)
Given (-1.5, 3), (6, -2), (-3, 4)
Let us Assume A(x1, y1) = (-1.5, 3)
Let us Assume B(x2, y2) = (6, -2)
Let us Assume C(x3, y3) = (-3, 4)
Area of triangle
Now,
Area of given triangle
= 0 square units
Find the area of the triangle whose vertices are
(-5, -1), (3, -5), (5, 2)
Given (-5, -1), (3, -5), (5, 2)
Let us Assume A(x1, y1) = (-5, -1)
Let us Assume B(x2, y2) = (3, -5)
Let us Assume C(x3, y3) = (5, 2)
Area of triangle
Now,
Area of given triangle
= 32 square units
Find the area of the triangle whose vertices are
(5, 2), (4, 7), (7, -4)
Given (5, 2), (4, 7), (7, -4)
Let us Assume A(x1, y1) = (5, 2)
Let us Assume B(x2, y2) = (4, 7)
Let us Assume C(x3, y3) = (7, -4)
Area of triangle
Now,
Area of given triangle
= 2 square units
Find the area of the triangle whose vertices are
(2, 3), (-1, 0), (2, -4)
Given (2, 3), (-1, 0), (2, -4)
Let us Assume A(x1, y1) = (2, 3)
Let us Assume B(x2, y2) = (-1, 0)
Let us Assume C(x3, y3) = (2, -4)
Area of triangle
Now,
Area of given triangle
Square units
Find the area of the triangle whose vertices are
(1, -1), (-4, 6), (-3, -5)
Given (1, -1), (-4, 6), (-3, -5)
Let us Assume A(x1, y1) = (1, -1)
Let us Assume B(x2, y2) = (-4, 6)
Let us Assume C(x3, y3) = (-3, -5)
Area of triangle
Now,
Area of given triangle
Square units
Given
Let us Assume A(x1, y1) =
Let us Assume B(x2, y2) =
Let us Assume C(x3, y3) =
Area of triangle
Now,
Area of given triangle
Square units
Find the area of the triangle whose vertices are
(-5, 7), (-4, -5), (4, 5)
Given (-5, 7), (-4, -5), (4, 5)
Let us Assume A(x1, y1) = (-5, 7)
Let us Assume B(x2, y2) = (-4, -5)
Let us Assume C(x3, y3) = (4, 5)
Area of triangle
Now,
Area of given triangle
= 53 square units
Find the area of the quadrilateral whose vertices are
(1, 1), (7, -3), (12, 2) and (7, 21)
Given (1, 1), (7, -3), (12, 2) and (7, 21)
Let us Assume A(x1, y1) = (1, 1)
Let us Assume B(x2, y2) = (7, -3)
Let us Assume C(x3, y3) = (12, 2)
Let us Assume D(x4, y4) = (7, 21)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle
Then,
Area of given triangle ABC
= 25 square units
Area of given triangle ABC
= 107 square units
Area of quadrilateral ABCD = Area of ABC + Area of ACD
= 25 + 107
= 132 sq units.
Find the area of the quadrilateral whose vertices are
(-4, 5), (0, 7), (5, -5), and (-4, -2)
Given (-4, 5), (0, 7), (5, -5), and (-4, -2)
To Find: Find the area of quadrilateral.
Let us Assume A(x1, y1) = (-4, 5)
Let us Assume B(x2, y2) = (0, 7)
Let us Assume C(x3, y3) = (5, -5)
Let us Assume D(x4, y4) = (4, -2)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle
Then,
Area of given triangle ABC
Area of given triangle ABC
square units
Area of quadrilateral ABCD = Area of ABC + Area of ACD
Hence, Area of Quadrilateral ABCD sq units.
Find the area of the quadrilateral whose vertices are
Given (-5, 7), (-4, -5), (-1, -6) and (4, 5)
To Find: Find the area of quadrilateral.
Let us Assume A(x1, y1) = (-5, 7)
Let us Assume B(x2, y2) = (-4, -5)
Let us Assume C(x3, y3) = (-1, -6)
Let us Assume D(x4, y4) = (4, 5)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle
Then,
Area of given triangle ABC
Area of given triangle ABC
= 56 square units
Area of quadrilateral ABCD = Area of ABC + Area of ACD
Hence, Area sq units.
Find the area of the quadrilateral whose vertices are
Given (0, 0), (6, 0), (4, 3), and (0, 3)
To Find: Find the area of quadrilateral.
Let us Assume A(x1, y1) = (0, 0)
Let us Assume B(x2, y2) = (6, 0)
Let us Assume C(x3, y3) = (4, 3)
Let us Assume D(x4, y4) = (0, 3)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle
Then,
Area of given triangle ABC
= 9 Square units
Area of given triangle ABC
= 6 square units
Area of quadrilateral ABCD = Area of ABC + Area of ACD
= 9 + 6
Hence, Area = 15 sq units.
Find the area of the quadrilateral whose vertices are
Given (1, 0), (5, 3), (2, 7) and (-2, 4)
To Find: Find the area of the quadrilateral.
Let us Assume A(x1, y1) = (1, 0)
Let us Assume B(x2, y2) = (5, 3)
Let us Assume C(x3, y3) = (2, 7)
Let us Assume D(x4, y4) = (-2, 4)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle
Then,
Area of given triangle ABC
Square units
Area of given triangle ABC
square units
Area of quadrilateral ABCD = Area of ABC + Area of ACD
Hence, Area = 25 sq units.
Find the area of the quadrilateral whose vertices taken in order are (- 4, -2), (-3, -5), (3, -2) and (2, 3).
Given: The vertices of the quadrilateral be A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3).
Let join AC to form two triangles,
Now, We know that
Area of triangle
Then,
Area of triangle ABC
Square Units
Now, Area of triangle ACD
Square Units
Area of quadrilateral ABCD = Area of triangle ABC+ Area of triangle ACD
Hence, Area of quadrilateral ABCD = 28 square Units
A median of a triangle divides it into two triangles of equal area. Verify this result for ABC whose vertices are A(1, 2), B(2, 5), C(3, 1).
Given a triangle whose vertices A(1, 2), B(2, 5), C(3, 1)
Let AD is the median on side BC
D will be the mid-point of segment BC. Therefore,
Coordinate of D
Area of triangle
Then,
Area of triangle ABD
sq units
Area of triangle ACD
sq units
Hence, ∆ABD = ∆ACD
If A, B, C are the points (-1, 5), (3, 1), (5, 7) respectively and D, E, F are the middle points of BC, CA and AB respectively, prove that ΔABC = 4ΔDEF.
Given: ABC is a triangle with points (-1, 5), (3, 1), (5, 7)
To Find ABC=4DEF
We know that
Area of triangle
Then,
Area of triangle ABC
= 16
Now we have to find point D, E, and F.
Hence D is the midpoint of side BC then,
Coordinates of D
= (4, 4 )
Hence E is the midpoint of side AC then,
Coordinates of E
= (2, 6)
Hence F is the midpoint of side AB then,
Coordinates of F
= (1, 3)
Area of triangle
Now Area of triangle DEF
= 4
Therefore Area of ABC= 4 Area of DEF.
Hence Proved.
Three vertices of a triangle are A(1, 2), B(-3, 6) and C(5, 4). If D, E, and C, respectively, show that the area of triangle ABC is four times the area of triangle DEF.
Given: ABC is a triangle with points (1, 2), (-3, 6), (5, 4)
To prove: The area of triangle ABC is four times the area of triangle DEF
We know that
Area of triangle
Then,
Area of triangle ABC
= 12
Now we have to find point D, E, F
Hence D is the midpoint of side BC then,
Coordinates of D
= (1, 5 )
Hence E is the midpoint of side AC then,
Coordinates of E
= (3, 3)
Hence F is the midpoint of side AB then,
Coordinates of F
= (-1, 4)
Area of triangle
Now Area of triangle DEF
= 3
Therefore Area of ABC = 4 Area of DEF.
Hence, Proved.
Find the area of the triangle formed by joining the mid-points of the sides of the triangles whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Let ABC is a triangle with points (0, -1), (2, 1), (0, 3)
To Find: Ratio of area of triangle ABC to triangle DEF
We know that
Area of triangle
Then,
Area of triangle ABC
= 4
Now we have to find point D, E, and F.
Hence D is the midpoint of side BC then,
Coordinates of D
= (1, 2 )
Hence E is the midpoint of side AC then,
Coordinates of E
= (0, 1)
Hence F is the midpoint of side AB then,
Coordinates of F
= (1, 0)
Area of triangle
Now Area of triangle DEF
= 1
Therefore Area of ABC= 4 Area of DEF.
Then, The ratio of ∆DEF and ∆ABC = 1:4
Find the area of a triangle ABC if the coordinates of the middle points of the sides of the triangle are (-1, - 2), (6, 1) and (3, 5).
Given: Coordinates of middle points are D(-1, -2), E(6, 1) and F(3, 5).
To find: Area of triangle ABC
Area of triangle
Now Area of triangle DEF
square units
Hence the area of ABC is square units
The vertices of ΔABC are A(3, 0), B(0, 6) and (6, 9). A straight line DE divides AB and AC in the ratio 1:2 at D and E respectively, prove that
Given, ABC is a triangle with vertices A(3, 0), B(0, 6) and C (6, 9)
To find:
We know that
Area of triangle
Now Area of triangle DEF
square units
Now, According to the question,
DE internally divides AB in the ratio 1:2 hence
Coordinates of D
= (2, 2)
E internally divides AC in the ratio 1:2 hence
Coordinates of D
= (4, 3)
Now Area of triangle ADE
square units
Therefore, Area of ∆ABC sq. units
Hence, Area of ABC = 9. Area of ADE
If (t, t - 2), (t + 3, t) and (t + 2, t + 2) are the vertices of a triangle, show that its area is independent of t.
Given a triangle with vertices (t, t-2), (t+3, t) and (t+2, t+2)
Area of triangle
= 2 sq units
Hence, t is not dependant variable in the triangle.
If A(x, y), B(1, 2) and C(2, 1) are the vertices of a triangle of area 6 square unit, show that x+y=15 or-9
Given: A triangle with vertices A(x, y), B(1, 2) and C (2, 1)
To find: x + y = 15 or -9
The area is 6 square units.
Area of triangle
[x + 1 – y + 2y - 4] = 12
[x + y - 3] = 12
x + y =15
Hence, Proved
Prove that the points (a, b+c), (b, c+a) and (c, a+b) are collinear.
Given: A(a, b + c), B(b, c + a) and C(c, a + b)
To prove : Given points are collinear
We know the points are collinear if area(∆ABC)=0
Area of triangle
Then,
Area
= 0
Hence, Points are collinear.
If the points (x1y1), (x2, y2) and (x3, y3) be collinear, show that
Given : A(x1, y1), B(x2, y2) and C(x3, y3)
We know the points are collinear if area(∆ABC)=0
Area of triangle
Then, Area
Now, Divide by
Taking common
Hence, Proved.
If the points (a, b), (a1, b1) and (a-a1, b-b1) are collinear, show that
Given (a, b), (a1, b1) and (a-a1, b-b1)
We know the points are collinear if area(∆ABC)=0
Area of triangle
Then, Area
{ab + a1 b1} = ab – ab1 – a1 b + a1 b1
- ab1 - a1b = 0
-ab1 = a1b
Therofe, We can write as
Hence, Proved.
Show that the point (a, 0), (0, b) and (1, 1) are collinear if
Given: (a, 0), (0, b) and (1, 1)
We know the points are collinear if area(∆ABC)=0
Area of triangle
Then, Area
ab – a – b = 0
ab = a + b
Since,
Then, a + b = ab
Hence, Proved.
Find the values of x if the points (2x, 2x), (3, 2x+1) and (1, 0) are collinear.
Given (2x, 2x), (3, 2x+1) and (1, 0)
We know the points are collinear if the area(∆ABC)=0
Area of triangle
Then, Area
4x2 - 4x – 1 = 0
4x2 - 2x - 2x – 1 = 0
2x(2x - 1) - 1(2x - 1) = 0
(2x-1)(2x-1)=0
Hence,
Find the value of K if the points A(2, 3), B(4, k) and C(6, -3) are collinear.
Given A(2, 3), B(4, k) and C(6, -3) are collinear
To find: Find the value of K
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle =
Then,
= 2k+6 - 24 +18 - 6k = 0
= -4k = 0
Hence, K = 0
Find the value of K for which the points (7, -2), (5, 1), (3, k) are collinear.
Given A(7, -2), B(5, 1) and C(3, k) are collinear
To find: Find the value of K
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle =
Then,
= 7 – 7k +5k+10-9 = 0
= -2k + 8 = 0
Hence, K = 4
Find the value of K for which the points (8, 1), (k, -4), (2, -5) are collinear?
Given A(8, 1), B(k, -4) and C(2, -5) are collinear
To Find: Find the value of k
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle =
Then,
= 8(1)+k(-6)+2(3) = 0
= 8 - 6k + 6 = 0
= -6k = -14
Hence, K =
Find the value of P are the points (2, 1), (p, -1) and (-1, 3)collinear?
Given A(2, 1), B(p, -1) and C(-1, 3) are collinear
To find: Find the value of p
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle =
Then,
= 2(-4)+p(2)-2=0
= -8 +2p -2 = 0
= 2p = 10
Hence, p =
Show that the straight line joining the points A(0, -1) and B(15, 2) divides the line joining the points C(-1, 2)and D(4, -5) internally in the ratio 2:3.
Given, A(0, -1) B (15, 2) divides the line on points C(-1, 2) and D(4, -5)
To Prove. Straight line divides in the ratio 2:3 internally
The equation of line
Now, Equation of line BC
⇒ 5y + 5 = x
Therefore, x – 5y = 5 ---(1)
Now, Equation of line BC
⇒ 5(y - 2) = -7(x + 1)
⇒ 5y - 10 = -7x - 7
Therefore, 7x +5y = 3 ---(2)
On solving equation (1) and (2)
X = 1 y =
Now, Point of the intersection of AB and CD is O (1,
Let us Assume that AB divides CD at O in the ratio m:n, then
x coordinate of O =
1 =
= 4m – n = m+n
= 4m – m = n+n
= 3m = 2n
= Hence Proved
Find the area of the triangle whose vertices are
((a+1) (a+2), (a+2)), ((a+2) (a+3), (a+3)) and ((a, 3) (a+4), (a+4))
Given, A triangle whose vertices are A ((a+1)(a+2), (a+2))
B ((a+2)(a+3), (a+3)) and C = ((a+b)(a+4), (a+4)).
To find: Find the area of a triangle.
Since, Area of triangle =
Then, =
=
=
Common terms will be canceled out
=
Hence, = 1 sq unit
The point A divides the join of P(-5, 1) and Q(3, 5) in the ratio k:1. Find the two values of k for which the area of ΔABC, where B is (1, 5) and C is (7, - 2) is equal to 2 units in magnitude.
Given: A divides the join of P(-5, 1) and Q(3, 5) in the ratio k:1
To Find: Two values of K
A divides join of PQ in the ratio k:1 hence
Coordinates of A
Now, We have A , B (1, 5), and C(7, -2)
Now, The area of ABC is equal to the magnitude 2 (Given)
Area of ABC
14k - 66 = 4k + 4
14k - 66 = -4k - 4
10k = 70
18k = 62
Hence, k = 7 and
The coordinates of A, B, C, D are (6, 3), (-3, 5), (4, -2) and (x, 3x) respectively. If , find x.
Given A, B, C, and D are (6, 3), (-3, 5), (4, -2) and (x, 3x) respectively.
and ∆DBC = 2∆ABC
To find: Find x.
Since Area of triangle
Now, The area of ∆DBC
= 11x – 7 sq units
Now, The area of ∆DBC
According to question, ∆DBC = 2∆ABC
⇒ 49 = 4(11x – 7)
⇒ 49 = 44x – 28
⇒ 44x = 77
Hence,
If the area of the quadrilateral whose angular points taken in order are (1, 2), (-5, 6), (7, -4) and (h, -2) be zero, show that h=3.
Given: vertices of the quadrilateral be A(1, 2), B(-5, 6), C(7, -4) and D(h, -2).
Let join AC to form two triangles,
Now, We know that
Area of triangle =
Then,
Area of triangle ABC =
=
=
= 6 sq units
Now, Area of triangle ADC =
=
=
= 3h - 15
Area of quadrilateral ABCD = Area of triangle ABC+ Area of triangle ADC
= 3h – 15 + 6
= 3h = 9
= h = 3
Hence, h is 3
Find the area of the triangle whose vertices A, B, C are (3, 4) (-4, 3), (8, 6) respectively and hence find the length of the perpendicular from A to BC.
Given: A triangle whose vertices A (3, 4) B(-4, 3), C(8, 6)
To find: Find the area of Triangle and length of AD
Area of triangle =
Then,
Area of triangle ABC =
=
=
= 4 square units
We need to find the length of AD on BC
Hence, We need to find the slope first,
The slope
Now the slope of BC = = 5
If AD perpendicular BC then the slope of AD is
Therefore, The equation of is (y - y1) = m(x - x1)
(y + 1) = (x - 5)
5y + 5 = - x + 5
The coordinates of the centroid of a triangle and those of two of its vertices are A(3, 1), B(1, -3) and the centroid of the triangle lies on the x-axis. Find the coordinates of the third vertex C.
Given: A(3, 1) and B(1, -3)
To find: Find the coordinate of the third vertex C.
Let Assume C = (a, b)
Centroid on C
Therefore, G(1, 0) as it lies on x-axis
⇒ 4 + a = 3
⇒ a = -1
⇒ - 2 + b = 0
⇒ b = 2
Hence, C is (-1, 2)
The area of a triangle is 3 square units. Two of its vertices are A(3,1), B(1,-3) and the centroid of the triangle lies on x-axis. Find the coordinates of the third vertex C.
Given: Coordinates of Triangle are A(3, 1) and B(1, -3)
Centroid of triangle lies on x- axis.
Let the third coordinate be C(x, y)
Centroid of the triangle with vertices (x1, y1), (x2, y2), (x3, y3) is given by,
C(X, Y)
The y coordinate of centroid will be 0, as it lies opn x – axis.
Therefore,
y1 + y2 + y3 = 0
1 – 3 + y3 = 0
y3 = 2
So, C(x, y) becomes (x, 2)
We are given that the area of triangle = 3 square units.
We know that,
Area of triangle
Putting the values we get,
3
-15 – 1 + 4x = 6
4x = 22
Hence, coordinates of the third vertex are
The area of a parallelogram is 12 square units. Two of its vertices are the points A(-1, 3) and B (-2, 4). Find the other two vertices of the parallelogram, if the point of intersection of diagonals lies on x-axis on its positive side.
Given: The area of a parallelogram is 12. A(-1, 3) and B(-2, 4)
To find: Find the other two vertices of the parallelogram.
Let C is (x, y) and A(-1, 3)
Since, AC is bisected at P, y coordinate ( when p = 0)
Then,
y = -3
So, Coordinate of C is (x, -3)
Now, Area of parallelogram ABCD = area of triangle ABC + Area of triangle BAD
Since, 2(Area of triangles) = area of parallelogram
We have, A(-1, 3) B(-2, 4) and C(x, -3)
Now, Area of triangle
Then,
Area of triangle ABC
6
6
12 = 5 - x
So, x = -7
Hence, Coordinate of C is (-7, -3)
In the same we will calculate for D
Let D is (x, y) and A(-2, 4)
Since, BD is bisected at Q, y coordinate ( when Q = 0)
Then,
y = -4
So, Coordinate of C is (x, -4)
We have, A(-1, 3) B(-2, 4) and C(x, -4)
Now, Area of triangle
Then,
Area of triangle ABC
6
6
12 = 6 - x
So, x = -6
Hence, Coordinate of D is (-6, -4)
Hence, C (-7, -3) and D(-6, -4)
Prove that the quadrilateral whose vertices are A(-2, 5), B(4, -1), C(9, 1) and D(3, 7) is a parallelogram and find its area. If E divides AC in the ratio 2:1, prove that D, E and the middle point F of BC are collinear.
Given: Let ABCD is a quadrilateral whose vertices A(-2, 5), B(4, -1), C(9, 1) and D(3, 7).
To prove: ABCD is a parallelogram .
We have to find |AD|, |AB|, |BC|, |DC|
The distance between two sides
|AD|
= √29
|AB| =
= √72
|DC| =
= √72
|BC| =
= √29
Therefore, AB = DC and AD = BC
Hence, ABCD is a parallelogram
Now, The Area of ABCD is = |a×b| =
=
= 0i – 0j+ 42 k
|a×b| = 42
Hence The area of parallelgram is 42
Prove that points (-3, -1), (2, -1), (1, 1) and (-2, 1) taken in order are the vertices of a trapezium.
Given: Points of the quadrilatreral A(-3, -1), B(2, -1), C(1, 1), and D(-2, 1).
To Prove: ABCD is a trapezium
Proof:
For proving ABCD to be a trapezium, we need to prove that two of the sides are parallel.
Therefore, AB and CD are parallel.
For proving ABCD a trapezium,
Slope of AB = Slope of CD
Slope of AB
Slope of CD
Hence, the quadrilateral is a trapezium.