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Co-ordinate Geometry: Distance Formula

Class 9th Mathematics West Bengal Board Solution
Let Us Work 4
  1. Let us calculate the distances of the following points from origin: (i) (7, -24) (ii)…
  2. Let us calculate the distances between the two points given below: (i) (5, 7) and (8,…
  3. Let us prove that the point (-2, -11) is equidistant from the two points (-3, 7) and…
  4. Let us show that the points (7, 9), (3, -7) and (-3, 3) are the vertices of a right…
  5. Let us prove that in both of the following cases, the three points are the vertices of…
  6. Let us prove that the three points A(3, 3), B(8, -2) and C(-2, -2) are the vertices of…
  7. Let us show by calculation that the points (2, 1), (0, 0), (-1, 2) and (1, 3) are the…
  8. Let us calculate and see that for what value of y, the distance of the two points (2,…
  9. Let us find a point on x-axis which is equidistant from the two points (3, 5) and (1,…
  10. Let us write by calculation whether the three points O (0, 0), A(4, 3) and B(8, 6) are…
  11. Let us show that the three points (2, 2), (-2, -2) and (- 2 root 3 , 2 root 3) are the…
  12. Let us show that the points (-7, 12), (19, 18), (15, -6) and (-11, -12) form a…
  13. Let us show that the points (2, -2), (8, 4), (5, 7) and (-1, 1) are the vertices of a…
  14. Let us show that the points (2, 5), (5, 9), (9, 12) and (6, 8) form a rhombus when…
  15. The distance between the two points (a + b, c - d) and (a - b, c + d) isA. 2 root a^2…
  16. If the distance between the two points (x, -7) and (3, -3) is 5 units, then the…
  17. If the distance of the point (x, 4) from the origin is 5 units, then the values of x…
  18. The triangle formed by the points (3, 0), (-3, 0) and (0, 3) isA. Equilateral B.…
  19. The co - ordinates of the center of the circle are (0, 0) and the co - ordinates of a…
  20. Short answer type questions: (i) Let us write the value of y if the distance of the…
Let Us Work 4
Question 1.

Let us calculate the distances of the following points from origin:

(i) (7, –24)

(ii) (3, –4)

(iii) (a + b, a – b)
Answer:

The distance of a point (x,y) from origin is given as


(i) Distance of a point (7,-24) from origin units


units


= 25 units


(ii) Distance of a point (3,-4) from origin


units


= 5 units


(iii) Distance of a point (a+b, a-b) from origin





Question 2.

Let us calculate the distances between the two points given below:

(i) (5, 7) and (8, 3)

(ii) (7, 0) and (2, –12)

(iii) and (0, –2)

(iv) (3, 6) and (–2, –6)

(v) (1, –3) and (8, 3)

(vi) (5, 7) and (8, 3)
Answer:

We know that distance between two points (x1, y1) and (x2, y2) is given by .


(i) Distance between the points (5, 7) and (8, 3)




= 5 units


(ii) Distance between the points (7, 0) and (2, -12)




= 13 units


(iii) Distance between the points and (0, -2)





units


(iv) Distance between the points (3, 6) and (–2, –6)




= 13 units


(v) Distance between the points (1, -3) and (8, 3)




= 9.21 units


(vi) Distance between the points (5, 7) and (8, 3)




= 5 units
Question 3.

Let us prove that the point (–2, –11) is equidistant from the two points (–3, 7) and (4, 6).
Answer:

We know that distance between two points (x1, y1) and (x2, y2) is given by .


Using the above distance formula,


Distance between (–2, –11) and (-3, 7)





Distance between (–2, –11) and (4, 6)





Hence proved that the point (–2, –11) is equidistant from the two points (–3, 7) and (4, 6).
Question 4.

Let us show that the points (7, 9), (3, –7) and (–3, 3) are the vertices of a right angled triangle by calculation.
Answer:

Let A → (7, 9)


B → (3, -7)


C → (-3, 3)


be the vertices of a triangle.


We know that distance between two points (x1, y1) and (x2, y2) is given by .


Using the above distance formula,


AB





BC





CA





Now, (BC)2 + (CA)2


and (AB)2


We find that (BC)2 + (CA)2 = (AB)2


Hence the points (7, 9), (3, –7) and (–3, 3) are the vertices of a right angled triangle.
Question 5.

Let us prove that in both of the following cases, the three points are the vertices of an isosceles triangle:

(i) (1, 4), (4, 1) and (8, 8)

(ii) (–2, –2), (2, 2) and (4, –4)
Answer:

(i) Let A → (1, 4)


B → (4, 1)


C → (8, 8)


be the vertices of a triangle.


We know that distance between two points (x1, y1) and (x2, y2) is given by .


Using the above distance formula,


AB





BC





CA





We find that BC = CA.


Since two sides of this triangle are equal, the points (7, 9), (3, –7) and (–3, 3) are the vertices of an isosceles triangle.


(ii) Let A → (-2, -2)


B → (2, 2)


C → (4, -4)


be the vertices of a triangle.


We know that distance between two points (x1, y1) and (x2, y2) is given by .


Using the above distance formula,


AB





BC





CA





We find that BC = CA.


Since two sides of this triangle are equal, the points (–2, –2), (2, 2) and (4, –4) are the vertices of an isosceles triangle.
Question 6.

Let us prove that the three points A(3, 3), B(8, –2) and C(–2, –2) are the vertices of a right-angled isosceles triangle. Let us calculate the length of the hypotenuse of ΔABC.
Answer:

Given A → (3, 3)


B → (8, -2)


C → (-2, -2)


are the vertices of a triangle.


We know that distance between two points (x1, y1) and (x2, y2) is given by .


Using the above distance formula,


AB





BC




= 10


CA





Now, (AB)2 + (CA)2


and (AB)2


We find that (AB)2 + (CA)2 = (BC)2 and AB = CA.


Hence proved that the three points A(3, 3), B(8, –2) and C(–2, –2) are the vertices of a right-angled isosceles triangle.


Length of hypotenuse BC = 10 units.
Question 7.

Let us show by calculation that the points (2, 1), (0, 0), (–1, 2) and (1, 3) are the angular points of a square.
Answer:

We have A → (2, 1)


B → (0, 0)


C → (-1, 2)


D → (1, 3)


We know that distance between two points (x1, y1) and (x2, y2) is given by .


Using the above distance formula,


AB





BC





CD





DA





Now, consider the diagonals, AC and BD.


AC





BD





We find that AB = BC = CD = DA with equal diagonals (AC = BD)


⇒ ABCD is a square.
Question 8.

Let us calculate and see that for what value of y, the distance of the two points (2, y) and (10, –9) will be 10.
Answer:

We know that distance between two points (x1, y1) and (x2, y2) is given by .


Using the above distance formula,


Distance between (2, y) and (10, -9)




Given, distance of the two points (2, y) and (10, –9) = 10.


∴


Squaring both sides, we get


y2 – 18y + 145 = 100


y2 – 18y + 45 = 0


On splitting the middle term,


y2 – 3y- 15y + 45 = 0


y(y – 3) - 15(y – 3) = 0


(y – 15) (y – 3) = 0


So, y can take two values, 15 and 3.
Question 9.

Let us find a point on x-axis which is equidistant from the two points (3, 5) and (1, 3).
Answer:

Let (x, 0) be a point on the x-axis equidistant from the two points (3, 5) and (1, 3).




Squaring both sides


x2 – 6x + 34 = x2 – 2x + 10


4x = 24


∴ x = 6


The required point is (6, 0).
Question 10.

Let us write by calculation whether the three points O (0, 0), A(4, 3) and B(8, 6) are collinear.
Answer:

We have O → (0, 0)


A → (4, 3)


B → (8, 6)


The distance of a point (x,y) from origin is given as , so


OA




= 5


and OB




= 10


We know that distance between two points (x1, y1) and (x2, y2) is given by .


Using the above distance formula,


AB





= 5


We find that,


OA + AB = 5 + 5 = 10 = OB


⇒ The points A, B and C are collinear.
Question 11.

Let us show that the three points (2, 2), (–2, –2) and are the vertices of an equilateral triangle.
Answer:


Let us mark the points on a graph sheet and call them A, B, C and join the three points to form a triangle.


Using distance formula,


Let us calculate the distance between A-B, B-C and A-C.


The distance between two points, D =


AB =



BC =




CA =



AB=BC=CA


Hence, Triangle ABC is equilateral.
Question 12.

Let us show that the points (–7, 12), (19, 18), (15, –6) and (–11, –12) form a parallelogram when they are joined orderly.
Answer:

Let us mark the points on a graph sheet and call them A, B, C, D and join the four points in the same order to form a four-sided figure.


A = (- 7, 12)


B = (19, 18)


C= (15, - 6)


D= (- 11, - 12)


We know that slope of a line is given by


Slopes of lines AB, BC, CD and DA are calculated as given below.


Slope (AB)


Slope (BC)


Slope (CD)


Slope (DA)


We see that slope of AB = Slope of CD


And


Slope of BC = Slope of DA


So AB ∥ CD and BC ∥ DA


ABCD is a parallelogram.
Question 13.

Let us show that the points (2, –2), (8, 4), (5, 7) and (–1, 1) are the vertices of a rectangle.
Answer:


Let


A = (2, - 2)


B = (8, 4)


C = (5, 7)


D = (- 1, 1)


Distance between two points =


AB


BC


CD


DA


Let’s join A and C, and B and D.


AC


BD = =


Since AB = CD, BC=AD and AC=BD, it is a rectangle.


(Opposite sides are equal; diagonals are equal)
Question 14.

Let us show that the points (2, 5), (5, 9), (9, 12) and (6, 8) form a rhombus when they are joined orderly.
Answer:

Let


A = (2, 5)


B = (5, 9)


C = (9, 12)


D = (6, 8)


Distance between two points =


AB = =


BC = = 5


CD = =


DA = =


Since all the sides are equal, the polygon qualifies as a rhombus.


The rhombus formed will look like:
Question 15.

The distance between the two points (a + b, c – d) and (a – b, c + d) is
A.

B.

C.

D.


Answer:

Distance between two points =


=



=
Question 16.

If the distance between the two points (x, –7) and (3, –3) is 5 units, then the values of x are
A. 0 or 3

B. 2 or 3

C. 5 or 1

D. –6 or 0
Answer:

Distance between two points =



Squaring on both sides,


25 = (3 - x) 2 + (- 3 + 7) 2


25 = x2 - 3x + 25


Or


x2 - 3x = 0


=x(x - 3) = 0


x=0 or x = 3
Question 17.

If the distance of the point (x, 4) from the origin is 5 units, then the values of x are
A. ± 4

B. ±

C. ± 3

D. none of these
Answer:

Distance between two points =


Here the second point is the origin.



Squaring on both sides,


25 = x2 + 16


Or


x= + 3 or - 3
Question 18.

The triangle formed by the points (3, 0), (–3, 0) and (0, 3) is
A. Equilateral

B. Isosceles

C. Scalene

D. Isosceles right-angled
Answer:

Here we need to calculate each side of the triangle.


Considering, A=(3,0), B=(0,3) and C=( - 3,0)


We have,





Since the two sides AB and BC are equal, so the triangle is isosceles.


Refer figure.
Question 19.

The co - ordinates of the center of the circle are (0, 0) and the co - ordinates of a point on the circumference are (3, 4), the length of the circle is
A. 5 units

B. 4 units

C. 3 units

D. None of these
Answer:

The length of the circle or radius is simply the distance between the two points.


The distance between two points =


Here the first point is the origin.


R =


= 5 units
Question 20.

Short answer type questions:

(i) Let us write the value of y if the distance of the point (–4, y) from the origin is 5 units.

(ii) Let us write the co - ordinates of a point on y - axis which is equidistant from two points (2, 3) and (–1, 2).

(iii) Let us write the coordinates of two points on the x-axis and y-axis for which an isosceles right-angled triangle is formed with the x-axis, y-axis and the straight line joining the two points.

(iv) Let us write the co - ordinates of two points on opposite sides of x - axis which are equidistant from x - axis.

(v) Let us write the co - ordinates of two points on opposite sides of y - axis which are equidistant from y - axis.
Answer:

(i) Distance between two points =

Here the second point is origin.


5 =


Squaring on both sides,


25 = 16 + y2


Or y = + 3 or - 3.


(ii) A point on y-axis will have x coordinate 0.


So, let the point on y-axis be (0, y) .


If this point is equidistant from given two points,


=


Squaring on both sides,


(y - 3) 2 + 4 = (y - 2) 2 + 1


Or y2 - 3y + 9 + 4 = y2 - 2y + 4 + 1


13 - 5 = y


y = 8


So the coordinate on the y-axis is (0, 8)


(iii) Let the point on x-axis be A (x, 0)


Let the point on y-axis be B (0, y)


To satisfy the conditions given in the problem,


The distance of A from origin should be the same as the distance of B from origin.


The distance between two points =



Or x= + y or x = - y


The coordinates should be (0, x) and (x, 0) where x is any real number.


(iv) (x, 0) and (- x, 0)


(v) (0, y) and (0, - y)