Question

. A point P is equidistant from R(-2, 4) and S(6,-4)

and its x-coordinate is twice its y-coordinate.
(0) Find the coordinates of P.
(ii) Hence, determine whether P, R and S are
collinear, showing your working clearly.​

Answer 1

I) coordinate of P is (11, 5.5)

II) Points are not collinear

Explanation:

I) Let the coordinates of P be (a, b)

We are told that its x-coordinate is twice its y-coordinate. Thus, a = 2b

Now, P is equidistant from R(-2, 4) and S(6,-4). Thus;

PR = PS

Using the formula for distance between two coordinates, we have;

√((a - (-2))² + (b - 4)²) = √((a - 6)² + (b - (-4))²)

This gives;

(a + 2)² + (b - 4)² = (a - 6)² + (b + 4)²

Since a = 2b,then we have;

(2b + 2)² + (b - 4)² = (2b - 6)² + (b + 4)²

4b² + 8b + 4 + b² - 8b + 16 = 4b² - 12b + 36 + b² + 8b + 16

Simplifying this gives;

20 = 42 - 4b

42 - 20 = 4b

4b = 22

b = 22/4

b = 11/2

b = 5.5

Since a = 2b

Then a = 2 × 5.5

a = 11

Thus, coordinate of P is (11, 5.5)

II) Formula for area of triangle formed by three points is given as;

A = ½|(x1 - x2) (x2 - x3)|

|(y1 - y2) (y2 - y3)|

x1 - x2 = 11 - (-2) = 13

x2 - x3 = -2 - 6 = -8

y1 - y2 = 5.5 - 4 = 1.5

y2 - y3 = 4 - (-4) = 8

Thus;

A = ½|13 -8|

|1.5 8|

A = ½((13 × 8) - (1.5 × - 8))

A = ½(104 + 12)

A = 58

Since the area is not zero, then it means the points are not collinear.