Cartesion System - Test Questions

Question 1

Question:   If the distance between the points (5, -6) and (x, -1) is 13, find the positive values of x. [3 Mark]

Answer:    The distance between two points is given by

Given A (5, -6), B (x, -1), AB = 13





Squaring both the sides,











Solving this quadratic equation,









Given x is a positive value, hence x = 17.

Question 2

Question:   Show that the point A (7, 7) is equidistant from the points B (3, 4) and C (4, 3). [3 Mark]

Answer:    Given, A (7, 7) is equidistant from B (3, 4) and C (4, 3)

Hence AB is given by distance formula,









= 5 units









= 5 units

Hence AB = AC

Question 3

Question:   Calculate the distance between the points P(3, 5), Q(6, 7). Leave your answer in root form. [2 Mark]

Answer:   

A(x₁, y₁), B(x₂, y₂) distance between AB by distance formula is,

Question 4

Question:   Calculate the distance between the points (-3, 5) and (4, -6), leave your answer in surd form. [2 Mark]

Answer:    Let A(-3, 5) and B(4, -6)

Question 5

Question:   Prove that the points A(-5, 4), B(-1, -2) and C(5, 2) are the vertices of an isosceles right angled triangle. [3 Mark]

Answer:    Given that ABC is a triangle.



If ABC is an isosceles D le, then AB = BC.













Hence ABC is an isosceles triangle.

If it is a right angled triangle then,

AC² = AB² + BC²







AC² = 104 units





= 2 x 52

= 104 units

Hence AC² = AB² + BC² and AB = BC

is an isosceles right angled triangle.

Question 6

Question:   Prove that the points (3, 1), (4, 3), (0, 3) and (-1, 1) are the vertices of a parallelogram. Or diagonals bisect each other. [3 Mark]

Answer:   

Let A(3, 1), B(4, 3), C(0, 3), D(-1, 1) the vertices of a (parallelogram) Quadrilateral.

In a parallelogram opposite sides are equal and parallel.

Hence if AB = CD and AD = BC, then ABCD is a parallelogram.

By distance formula,



















Hence one pair of opposite sides are equal.

Consider AD and BC,







= 4 units









= 4 units

Hence AB = DC and AD = BC

ABCD is a parallelogram.

Question 7

Question:   Prove that the triangle PQR with vertices P (5, -3), Q (5, 0), R (2, 0) is an isosceles triangle. [2 Mark]

Answer:   

If triangle PQR is an isosceles triangle then either PQ = QR,

(or any 2 sides are equal)



PQ









= 3 units







PQ = QR Hence is isosceles.

Question 8

Question:   Find the Co-ordinates of the points whose abscissa is 10 and which are at a distance of 10 units from (2, -3). [3 Mark]

Answer:    Let the point A whose abscissa is 10 have the co-ordinates as A (10, y).

Another point B (2, -3)

AB = 10 units (given)

According to distance formula,













Squaring both the sides,

100 = y² + 6y + 73



Solving the quadratic,

y² + 6y - 27 = 0

y² + 9y - 3y - 27 = 0

y(y + 9) - 3(y + 9) = 0

(y + 9(y - 3) = 0

y = -9 or y = 3

Vertices of ordinates are -9 or 3

Points are, (10, -9) or (10, 3)

Question 9

Question:   Prove that A (-4, 3), B (0, 9), C (-3, 11) and D (-7, 5) form a parallelogram ABCD. [3 Mark]

Answer:    If ABCD are the vertices of the parallelogram, then their opposite sides are equal in length and parallel.

AB = CD and BC = AD















Hence AB = CD















Hence BC = AD

ABCD is a parallelogram.

Question 10

Question:   Prove that P (-5, 4), Q (-1, -2), R (5, 2) and S (1, 8) form a square PQRS. [3 Mark]

Answer:   



According to the property of square, if PQRS is a square then all the sides are equalPR = QS































Hence PQ = QR = RS = SP.

All the sides are equal.

One of the diagonals,



















Hence PR = SQ

Diagonals are equal.

Hence PQRS is a square.