Long Questions (Question 6)

Question 7:

Solutions by scale drawing will not be accepted.

Diagram below shows a triangle OPQ. Point S lies on the line PQ.

(a) A point Y moves such that its distance from point S is always 5 uints.

Find the equation of the locus of Y.

(b) It is given that point P and point Q lie on the locus of Y .

Calculate

(i) the value of k,

(ii) the coordinates of Q.

(c) Hence, find the area, in uint², of triangle OPQ.

Solution:
(a)

\begin{array}{l} The equation of the locus Y (x, y) is given by Y S = 5 units \\ \sqrt{{(x - 5)}^{2} + {(y - 3)}^{2}} = 5 \\ x^{2} - 10 x + 25 + y^{2} - 6 y + 9 = 25 \\ x^{2} + y^{2} - 10 x - 6 y + 9 = 0 \end{array}

(b)(i)

Given P (2, k) lies on the locus of Y.

(2)² + (k)²– 10(2) – 6(k) + 9 = 0

4 + k²– 20 – 6k + 9 = 0

k² – 6k – 7 = 0

(k – 7) (k + 1) = 0

k = 7 or k = – 1
Based on the diagram, k = 7.

(b)(ii)

As P and Q lie on the locus of Y, S is the midpoint of PQ. P = (2, 7), S = (5, 3).

Let the coordinates of Q = (x, y),

\begin{array}{l} (\frac{2 + x}{2}, \frac{7 + y}{2}) = (5, 3) \\ \frac{2 + x}{2} = 5 and \frac{7 + y}{2} = 3 \\ 2 + x = 10 and       7 + y = 6 \\ x = 8 and y = - 1 \end{array}

Coordinates of point Q = (8, –1).

(c)

\begin{array}{l} Area of △ O P Q \\ = \frac{1}{2} | \begin{array}{l} 0 8     2 \\ 0 - 17 \end{array} \begin{array}{l} 0 \\ 0 \end{array} | \\ = \frac{1}{2} | 0 + (8) (7) + 0 - 0 - (- 1) (2) - 0 | \\ = \frac{1}{2} | 58 | \\ = 29 {units}^{2} \end{array}