Cartesion System - Test Questions-i

Question 1

Question:   Find the distance between the following pair of points.[2 Mark]

(i) (-5, 3), (3, 1)

(ii) (4, 5), (-3, 2)

Answer:    (i) Let A (x₁, y₁) = (-5, 3)

B (x₂, y₂) = (3, 1)

The distance between the points is



(ii) A = (4, 5) and B = (-3, 2)

Let (x₁, y₁) = (4,5) and (x₂, y₂) = (-3, 2)

The distance between the points is

Question 2

Question:   Find the distance between the following pairs of points.[Each 2 Marks]

(a) (-a, b) and (a, b)

Answer:    (a) (x₁, y₁) = (-a, b) (x₂, y₂) = (a, b)

The distance between the two points is





Distance between the two points is







Distance between the two points is









Distance between the two points is

Question 3

Question:   Use distance formula, show that the points (8, 7), (6, 4) and (0, -5) are collinear. [3 Mark]

Answer:    Let A = (8, 7), B = (6, 4), C = (0, -5)



Out of the line segments AB, BC and AC, if the sum of two of them is equal to the third, then A, B, C are collinear.

The three given points A, B, C are collinear.

Question 4

Question:   A point is equidistant from A (-6, 4) and B (2, -8). Find its co-ordinates, if its abscissa and ordinate are equal. [2 Mark]

Answer:   

Let P be the point which is equidistant from A and B. Since the abscissa and ordinates are equal, let (x, x) be the co-ordinates of P.

PA = PB (given)

Question 5

Question:   The distance between A (1, 3) and B (x, 7) is 5. Calculate the possible values of x. [2 Mark]

Answer:   





Squaring both sides, we get

25 = x²- 2x + 17

Question 6

Question:   What points on the x-axis are at a distance of 5 units from the point (5, -4)? [2 Mark]

Answer:    Note that all the points on x-axis have y-co-ordinate zero. Let (x, 0) be the point which is at a distance of 5 units from the point (5, -4).

Using distance formula, we have



Squaring both sides, we get

(5 - x)²+ 16 = 25



The points are (8, 0) and (2, 0).

Question 7

Question:   What point on y-axis is equidistant from the points (7, 6) and (-3, 4)? [2 Mark]

Answer:    Note that all the points on y-axis has x co-ordinate 0. Let P (0, y) be the point on Y axis which is equidistant from A (7,6) and B (-3,4).

PA = PB (given)

(7 - 0)² + (6 - y)²

(- 3 - 0)² + (4 - y)²

Squaring both sides and simplifying, we get

49 + 36 + y²- 12y = 9 + 16 - 8y + y²

-12y + 8y = 25 - 49 - 36

-4y = -60



The point (0, 15) is equidistant from the points (7,6) and (-3,4).

Question 8

Question:   A is a point on the y-axis whose ordinate is 5 and B is the point (-3, 1). Calculate the length of AB. [2 Mark]

Answer:    The co-ordinates of A are (0,5)

B = (-3,-1) (given)

Question 9

Question:   Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11. [2 Mark]

Answer:    The co-ordinates of B are (11, 0).

(on the x-axis, y = 0, abscissa = x-coordinate = 11)

Question 10

Question:   A point A is at a distance from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice of its abscissa. [3 Mark]

Answer:    Ordinate is y-co-ordinate and Abscissa is x co-ordinate.

Let the co-ordinates of A be (x, y).

Then, y = 2x (given)

A is a point which has co-ordinates (x, 2x) and B (4, 3).





The two possible points are (3, 6) and (1, 2) .