Coordinate Geometry

Co-ordinates of A, B, C, D and E are as follows,

A(-6 , 5 ) , B (5 , 4) , C(-3 , 2) , D (-2 , 2) , E(-1 , 4)

(i) (7, 0) lies on X-axis.

(ii) (0, -5) lies on negative or Y-axis.

(iii) (0, 1) lies on positive Y-axis.

(iv) (-4, 0) lies on negative X-axis.

(i) In given points X co-ordinate is negative and Y co-ordinate is positive, Hence, (-6, 5) lies in 2^nd quadrant.

(ii) In given points X co-ordinate is negative and Y co-ordinate is also negative, Hence, (-3, -2 ) lies in 3^rd quadrant.

(iii) In given points X co-ordinate is positive and Y co-ordinate is negative, Hence, (2, -9) lies in 4^th quadrant.

The given equation is y = x + 1 ……….(i)

Now,

By putting x = 0 in equation (i), we get y = 1

By putting x = 1 in equation (i), we get y = 2

By putting x = 2 in equation (i), we get y = 3

By putting x = 3 in equation (i), we get y = 4

A table is form such that:

The given equation is y = 3x + 2 …….(i)

Now,

By putting x = -1 in equation (i), we get y = -1

By putting x = 0 in equation (i), we get y = 2

By putting x = 1 in equation (i), we get y = 5

By putting x = 2 in equation (i), we get y = 8

A table is form such that:

The given equation is y = 5x - 3 …….(i)

Now,

By putting x = 0 in equation (i), we get y = -3

By putting x = 1 in equation (i), we get y = 2

By putting x = 2 in equation (i), we get y = 7

By putting x = 3 in equation (i), we get y = 12

A table is form such that :

The given equation is y = 3x …….(i)

Now,

By putting x = 0 in equation (i), we get y = 0

By putting x = 1 in equation (i), we get y = 3

By putting x = 2 in equation (i), we get y = 6

By putting x = 3 in equation (i), we get y = 9

A table is form such that:

The given equation is y = - x …….(i)

Now,

By putting x = -2 in equation (i), we get y = 2

By putting x = -1 in equation (i), we get y = 1

By putting x = 0 in equation (i), we get y = 0

By putting x = 1 in equation (i), we get y = -1

A table is form such that:

We can see that, x – coordinate is negative and y- coordinate is positive.

Hence, it can be clearly said that P (-5,3) lies in the 2^nd quadrant.

∴ Option B is correct

We can see that, x – coordinate is positive and y- coordinate is negative.

Hence, it can be clearly said that Q (4,-6) lies in the 4^th quadrant.

∴ Option D is correct.

We can see that, x – coordinate is zero and y- coordinate is negative.

Hence, it can be clearly said that Q (0,-4) lies on the y axis.

∴ Option D is correct.

We can see that, x – coordinate is positive and y- coordinate is zero.

Hence, it can be clearly said that B (8,0) lies on the x axis.

∴ Option C is correct.

We can see that, x – coordinate is negative and y- coordinate is zero.

Hence, it can be clearly said that C(-6,0) lies on the x axis.

∴ Option C is correct.

Origin is the point of intersection of the two coordinate axes.

∴ Option C is correct.

We have, x > 0 and y < 0

∴x is positive and y is negative

= point (x, -y) lies in 4^th quadrant

∴ Option D is correct.

We know that abscissa and ordinate can be equal in only two cases i.e.

i. x and y both are positive

ii. x and y both are negative

∴The points will lie in the 1^st and 3^rd quadrants only.

∴ Option C is correct.

We know that abscissa and ordinate can have different signs in only two cases i.e.

i. x is negative and y is positive

ii. x is positive and y is negative

∴The points will lie in the 2^nd and 4^th quadrants only.

∴ Option C is correct.

Since, we have the point A (7, 5)

And we have to find its perpendicular distance from y-axis

∴The perpendicular distance will be the x- coordinate

Hence, it is 7 units.

∴ Option A if correct.

We have, both the coordinates are negative i.e.

x and y both are negative

Hence, the point lies in the 3^rd quadrant.

∴ Option C is correct

We have, (x, y)

Where, x is positive

Hence it may lie in either 1^st or 4^th quadrant.

∴ Option D is correct.

Here, abscissa of A = 3

Abscissa of B = -2

According to the question,

(abscissa of A) – (abscissa of B) = 3 – (-2)

= 3 + 2 = 5

∴ Option C is correct.

Let us see the plot of these points,



Since, all the given points have their x – coordinate positive and y – coordinate negative.

Hence, all these points lie in the 4^th quadrant. And also they all lie in a straight line.

∴ Option C is correct

We know that, a point can only lie on x-axis if its y coordinate is 0.

Hence, points A, C and E does not lie on x-axis

∴ Option C is correct

Since, in the 2^nd quadrant x is negative and y is positive

Hence, as per sign the point could be written as (-, +)

∴ Option B is correct.

We have,

y = 3x + 4 (i)

a. (1, 7) [putting x = 1 and y = 7 in (i)]

7 = 3(1) + 4

7 = 7

Thus, A lies on the line.

b. (2, 10) [putting x = 2 and y = 10 in (i)]

10 = 3(2) + 4

10 = 10

Thus, B lies on the line.

c. (-1, 1) [putting x = -1 and y = 1 in (i)]

1 = 3(-1) + 4

1 = 1

Thus, C lie on the line.

d. (4, 12) [putting x = 4 and y = 12 in (i)]

12 = 3(4) + 4

12 ≠ 16

Thus, D does not lie on the line.

∴ Option D is correct

We have, y = 2x + 3 (i)

a. (2, 8) [putting x = 2 and y = 8 in (i)]

8 = 2(2) + 3

8 ≠ 7

Thus, A doesn’t lie on the line.

b. (3, 9) [putting x = 3 and y = 9 in (i)]

9 = 2(3) + 3

9 = 9

Thus, B lies on the line.

∴ Option B is correct.

We have, a < 0 and b < 0

i.e. both x and y are negative

Hence, point P lies in 3^rd quadrant

∴ Option C is correct.

Since, we have the point P (4, 3)

And we have to find its perpendicular distance from y-axis

∴The perpendicular distance will be the x- coordinate

Hence, it is 4 units.

∴ Option B is correct

Here, OA = 4 – 0 = 4 units

OB = 6 – 0 = 6 units

= 12 sq. units

∴ Option B is correct.

Here, OA = 4 – 0 = 4 units

OB = 6 – 0 = 6 units

= 12 sq. units

∴ Option B is correct

Here,

1^st statement is true as the point lying on x axis has its y coordinate as 0

2^nd statement is true as the point lying on y axis has its x coordinate as 0

3^rd statement is false as a point can never lie on both the axes unless it is their point of intersection i.e. (0, 0)

∴ Option A is correct

We have, pb(-3, 0)

Since, the y coordinate is 0.

Hence, it lies on x-axis.

∴ Option (a) is correct.

We have,

Point O(0, 0)

It does not lie in any of the quadrant because it is the point of intersection of both the axes.

∴ Option (d) is the correct option.

We know that,

The point P (-6, -4) lies in the third quadrant as the points of the type (-, -) lie in III quadrant

Also, we know that

The signs of points in quadrants I, II, III and IV are respectively (+,+) (-,+) (-,-) and (+,-)

∴ Both assertion and reason are true and reason justifies the assertion

Hence, option (a) is correct

According to question,

If a b then (a, b) (b, a)

This statement is true

Also, (4, -3) lies in the quadrant IV

As both assertion and reason are true but the reason does not justify the assertion

Hence, option (b) is correct

(i) The given statement is false as the ordinate of the point P (6, 0) is 0 and hence it lies on the x-axis

(ii) The given statement is also false because the perpendicular distance of the point A (5, 4) from the x-axis will be 4 units instead of 5 units

(i) The given statement is false because the mirror image of the point A (4, 5) on the x-axis is A’ (4, 5) instead of A’ (-4, 5)

(ii) The given statement is true as the mirror image of the point A (4, 5) in the y-axis is A’ (-4, 5)

A. The given statement is true as the ordinate of the point is 0 which lies on the x-axis

B. The given statement is false as the point (0, -3) lies on the y-axis

(a) We know that,

The points that lie on the x-axis have coordinate = 0

∴ The equation of the x-axis will be y = 0

(b) We know that,

The points that lie on the y-axis have abscissa = 0

∴ The equation of the y-axis will be x = 0

(c) We know that,

Any point on the x-axis is of the form (a, 0)

(d) We also know that,

A point on the y-axis is of the form = (0, b)

Hence the correct match for the given table is as follows:

(a) – (q)

(b) – (s)

(c) - (p)

(d) – (r)

(a) We know that,

The point of the type (a, 0) lies on the x-axis

∴ Point A (-3, 0) lies on the x-axis

(b) We know that,

The point of the type (-, -) lie in the III quadrant

∴ Point B (-5, -1) lies in the quadrant III

(c) We know that,

The point of the type (+, -) lie in the quadrant IV

∴ Point C (2, -3) lies in the quadrant IV

(d) We know that,

The point of the type (o, b) lies on the y-axis

∴ Point D (0, -6) lies on the y-axis

Hence, the correct match for the above given table is as follows:

(a) - (s)

(b) – (r)

(c) – (q)

(d) – (p)

A. We know that,

Point (-3, 6) lie in the second quadrant

B. We know that,

Point (4, -6) lie in the fourth quadrant

C. We know that,

Point (-5, -7) lie in the third quadrant

D. We know that,

Point (5, 3) lie in the first quadrant

The required point is shown in the graph given below:

In the above graph PL is drawn perpendicular to x-axis while PM is drawn perpendicular to y-axis

∴ Coordinates of L = (-6, 0)

Also, coordinates of M = (0, 3)

The given four points A (-5, 2), B (3, -2), C (-4, -3) and D (6, 0) are plotted on the graph paper as follows:

Let the vertices of the triangle be A (1, 4), B (-2, 2) and C (3, 2)

Now, when we plot and join these points on the graph paper, we get a triangle ABC

Let the line BC intersect y-axis at D

∴ BC = BD + DC

= (2 + 3) units

= 5 units

Now, AL is drawn perpendicular to x-axis meeting BC at L

∴ Ordinate of point L = Ordinate of point C – 2

AL = AM – LM

= 4 – 2

= 2 units

Hence, area of =

=

=

= 5 units

∴ Area of the triangle ABC = 5 square units

Let the three vertices of rectangle ABCD be A (2, 2), B (-3, 2) and C (-3, 5)

Now, on plotting these points on the graph paper and by joining the points we get:

The point A lies in the first quadrant while B and C lie in the second quadrant

Let us assume D be the fourth vertex of the rectangle

∴ Abscissa of D = Abscissa of A = 2

And, Ordinate of D = Ordinate of C = 3

Hence, Coordinate of the fourth vertex, D = (2, 5)

Let y-axis cut AB and CD at point L and M respectively

∴ AB = (BL + LA)

= (3 + 2)

= 5 units

Also, BC = 5 – 2

= 3 units

∴ Area of rectangle ABCD = BC × AB

= 3 × 5

= 15 square units

Hence, the area of the rectangle ABCD is 15 square units

Let the three vertices of rectangle ABCD be A (3, 2), B (-2, 2) and D (3, -3)

Now, on plotting these points on the graph paper and by joining the points we get:

A, B and D lie in different quadrants

Let us assume D be the fourth vertex of the rectangle

∴ Abscissa of C = Abscissa of B = - 2

And, Ordinate of C = Ordinate of D = - 3

Hence, Coordinate of the fourth vertex, C = (-2, -3)

Let y-axis cut AB and CD at point L and M respectively

∴ AB = (BL + LA)

= (2 + 3)

= 5 units

∴ Area of rectangle ABCD = AB × AB

= 5 × 5

= 25 square units

Hence, the area of the rectangle ABCD is 25 square units

(i) We have,

Abscissa of point D = 0

Ordinate of point D = -5

∴ Coordinates of point D = (0, -5)

(ii) From the given graph, we have:

Abscissa of point A = - 4

(iii) From the given graph, we have:

Coordinates of point E = (2, -3)

(iv) From the given graph, we have:

Coordinates of point C = (-3, -4)

(v) From the given graph, we have:

Ordinate of point E = - 3

(vi) From the given graph, we have:

Point B lies on x-axis

∴ Abscissa of point B = - 2

Ordinate of point B = 0

Hence coordinates of point B are (-2, 0)

(vii) From the given graph, we have:

Abscissa of point F = 5

Ordinate of point F = - 1

(viii) From the given graph, we have:

Coordinates of the origin = (0, 0)

According to question, we have

x < 0 and y > 0 then these points will lie in second quadrant

As, points of the type (-, +) lie on the second quadrant

∴ Option B is correct

From the given options in the question the point which does not lie in any quadrant is (0, 3)

∴ Option D is correct

When we plot the given points in the graph paper then,

is the right angle triangle, where

OB = Base = 6 units

Height of triangle = OA = 6 units

∴ Area of =

=

=

= 18 square units

∴ Option C is correct

We know that,

Any point which lies on the x-axis is of the form (x, 0) for x

Also, point which lies on the y-axis is of the form (0, y) for y

Hence, statement I and II are true

∴ Option A is correct

From the given four options, (-5, 5) does not satisfy the given equation:

3x = 2x – 5

We have,

R.H.S = 2 × (-5) – 5

= - 10 – 5

= - 15

Also, L.H.S = 3 × 5

= 15

Hence, the point (-5, 5) does not lie on the line 3y = 2x – 5

∴ Option D is correct

The following given points are plotted on the graph paper as follows:

We have,

2y = 3 – 5x

Now, by putting the value of x = - 1 in the given equation we get:

2y = 3 – 5 × (-1)

2y = 3 + 5

2y = 8

y = = 4

Hence, when x = - 1 then y = 4

We have,

Abscissa of point A (0, -4) = 0

∴ Point A lies on the y-axis

From the given point given in the question, we have

The abscissa and the ordinate of the point B (-3, -5) are negative and we know that those points lie on the III quadrant

∴ Point B lies in the third quadrant

We have,

Abscissa of point P (-2, -3) = - 2

We know that, distance cannot be negative

∴ The perpendicular distance of the given point from the y-axis is 2 units

The coordinate axes meet at the origin i.e., at point O (0, 0)

(i) The given statement is false as the given point lies on the x-axis

(ii) The given statement is also false as the point is P (-4, -3)

(iii) The given statement is also false as the point A (1, -1) lies in the quadrant IV and point B (-1, 1) lies in the quadrant II

(iv) The given statement is false as the coordinates of the point are (0, 3)

(v) The given statement is true as the point C (0, -5) lies on y-axis

(vi) The given statement is also true as the point O (0, 0) lies on x-axis as well as y-axis

The concept is:

(+x, +y) 1^st Quadrant

(-x, +y) 2^nd Quadrant

(-x, -y) 3^rd Quadrant

(+x, -y) 4^th Quadrant

The given points are plotted as follows:

(i)The given four points G (-3, 0) , H (-8, 0), Q (4, 0) and R (9, 0) lie on the x-axis

∴ Their ordinates are equal to O

(ii) The given four points L (0, -6) , K (0, -2), D (0, 3) and C (0, 7) lie on the x-axis

∴ Their abscissa are equal to O

(iii) The ordinates of points M (-1, 3) , J (-4, -3) and P (6, 3) are equal to -3

(iv) Points having abscissa equal to 2 is B (2, 4) and N (2, -1)

(v) From the given points, points E and F lie in quadrant II

∴ Coordinates of E = (-4, 4)

Coordinates of F = (-6, 2)

(vi) Coordinates of all those points having abscissa and ordinate same value are:

A (3, 3) and I (-2, -2)

(i) The mirror image of the point (2, 5) in the x-axis is (2, -5)

(ii) The mirror image of the point (3, 6) in the y-axis is (-3, 6)

(iii) According to question,

Point (a, b) lies in the second quadrant so a must be a negative number and b must be a positive number

∴ Point (b, a) lies in the fourth quadrant

The concept is:

(+x, +y) 1^st Quadrant

(-x, +y) 2^nd Quadrant

(-x, -y) 3^rd Quadrant

(+x, -y) 4^th Quadrant

(i) Points having ordinate = 4 and abscissa = - 3 lies in the quadrant II

(ii) Points having ordinate = - 5 and abscissa = 4 lies in the quadrant IV

(iii) Points having ordinate = - 2 and abscissa = - 1 lies in the quadrant III

(iv) Points having ordinate = 3 and abscissa = - 5 lies in the quadrant II

(v) Points having ordinate = 1 and abscissa = 2 lies in the quadrant I

(vi) Points having ordinate = - 4 and abscissa = 7 lies in the quadrant IV

From the points given in the question, we have

Points B (2, 0) and D (6, 0) have their ordinates = 0

∴ They lie on the point x-axis

And the point whose ordinate is not equal to zero does not lie on the x-axis

∴ Points A, C, E and F do not lie on the x –axis

Let the three vertices of rectangle ABCD be A (3, 1), B (-3, 1) and C (-3, 3)

Now, on plotting these points on the graph paper and by joining the points we get:

A lies in thye first quadrant while B and C lie in the second quadrant

Let us assume D be the fourth vertex of the rectangle

∴ Abscissa of D = Abscissa of A = 3

And, Ordinate of D = Ordinate of C = 3

Hence, Coordinate of the fourth vertex, D = (3, 3)

It is given in the question that,

OABC is a rectangle where O is the origin and OA = 5 units along x-axis

Also, AB = 3 units and B lies in the fourth quadrant

Now, coordinates of origin, O = (0, 0)

As point A lies on the x-axis

∴ Coordinate of point A = (5, 0)

Also, point B lies in the fourth quadrant

So, coordinate of point B will be negative

As the given width AB = 3 units

∴ Coordinates of point B = (5, -3)

Also, point C and O lies on the same line

Thus, abscissa of C = abscissa of O = 0

Similarly, Point C and B lies on the same altitude

Hence, both points have equal altitude

∴ Coordinates of C = (0, -3)

Hence, the coordinates of the vertices of the given rectangle are:

O (0, 0), A (5, 0), B (5, -3) and C (0, -3)

Let the three vertices of triangle ABC be:

A (2, 5), B (-2, 2) and C (4, 2)

Now, when we plot these points in the graph paper then we see that,

Point A and C lie in the quadrant I and point B lie in the II quadrant

Let the line BC intersect y-axis at point D

∴ BC = (BD + DC)

= (2 + 4) units

= 6 units

Now, we have to draw AM perpendicular to x-axis and intersect BC at L

∴ Ordinate of point L = Ordinate of point B - Ordinate of point C

AL = AM – LM

= 5 – 2

= 3 units

Hence, Area of triangle ABC =

=

=

= 9 square units

∴ Area of triangle ABC = 9 square units