7.8.4 Coordinate Geometry Long Question (Question 7 & 8)

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}

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

(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}

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

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}

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

Question 8:
Diagram below shows a quadrilateral ABCD. Point C lies on the y-axis.

The equation of a straight line AD is 2y = 5x – 21
(a) Find
(i) the equation of the straight line AB,
(ii) the coordinates of A,

(b) A point P moves such that its distance from point D is always 5 units.
Find the equation of the locus of P.

Solution:
(a)(i)

\begin{array}{l} 2 y = 5 x - 21 \\ y = \frac{5}{2} x - \frac{21}{2} \\ m_{A D} = \frac{5}{2} \\ m_{A B} \times m_{A D} = - 1 \\ m_{A B} \times \frac{5}{2} = - 1 \\ m_{A B} = - \frac{2}{5} \\ Equation of A B \\ y - y_{1} = m_{A B} (x - x_{1}) \\ y + 1 = - \frac{2}{5} (x + 2) \\ 5 y + 5 = - 2 x - 4 \\ 5 y = - 2 x - 9 \end{array}

$\begin{array}{l} 2 y = 5 x - 21 \\ y = \frac{5}{2} x - \frac{21}{2} \\ m_{A D} = \frac{5}{2} \\ m_{A B} \times m_{A D} = - 1 \\ m_{A B} \times \frac{5}{2} = - 1 \\ m_{A B} = - \frac{2}{5} \\ Equation of A B \\ y - y_{1} = m_{A B} (x - x_{1}) \\ y + 1 = - \frac{2}{5} (x + 2) \\ 5 y + 5 = - 2 x - 4 \\ 5 y = - 2 x - 9 \end{array}$

(a)(ii)

\begin{array}{l} 2 y = 5 x - 21 ………. (1) \\ 5 y = - 2 x - 9 ………. (2) \\ (1) \times 5 : 10 y = 25 x - 105 ………. (3) \\ (2) \times 2 : 10 y = - 4 x - 18 ………. (4) \\ (2) - (4) : 0 = 29 x - 87 \\ x = 3 \\ From (1), \\ 2 y = 15 - 21 \\ 2 y = - 6 \\ y = - 3 \\ A = (3, - 3) \end{array}

$\begin{array}{l} 2 y = 5 x - 21 ………. (1) \\ 5 y = - 2 x - 9 ………. (2) \\ (1) \times 5 : 10 y = 25 x - 105 ………. (3) \\ (2) \times 2 : 10 y = - 4 x - 18 ………. (4) \\ (2) - (4) : 0 = 29 x - 87 \\ x = 3 \\ From (1), \\ 2 y = 15 - 21 \\ 2 y = - 6 \\ y = - 3 \\ A = (3, - 3) \end{array}$

(b)

\begin{array}{l} y = 2, \\ 4 = 5 x - 21 \\ 5 x = 25 \\ x = 5 \\ Point D = (5, 2) \\ P D = 5 \\ \sqrt{{(x - 5)}^{2} + {(y - 2)}^{2}} = 5 \\ {(x - 5)}^{2} + {(y - 2)}^{2} = 25 \\ x^{2} - 10 x + 25 + (y^{2} - 4 y + 4) = 25 \\ x^{2} + y^{2} - 10 x - 4 y + 4 = 0 \end{array}

$\begin{array}{l} y = 2, \\ 4 = 5 x - 21 \\ 5 x = 25 \\ x = 5 \\ Point D = (5, 2) \\ P D = 5 \\ \sqrt{{(x - 5)}^{2} + {(y - 2)}^{2}} = 5 \\ {(x - 5)}^{2} + {(y - 2)}^{2} = 25 \\ x^{2} - 10 x + 25 + (y^{2} - 4 y + 4) = 25 \\ x^{2} + y^{2} - 10 x - 4 y + 4 = 0 \end{array}$

2 thoughts on “7.8.4 Coordinate Geometry Long Question (Question 7 & 8)”

tye

January 13, 2021 at 10:09 am | Reply

for 7b (ii), how do i know that S is the midpoint of PQ? its not given in the question thanks
- myhometuition
  
  January 16, 2021 at 1:59 pm | Reply
  
  Hi Tye,
  It is given in the question that point P and point Q lie on the locus of Y.
  PSQ is a straight line and SP = SQ = 5 units, therefore S is the midpoint of PQ.

2 thoughts on “7.8.4 Coordinate Geometry Long Question (Question 7 & 8)”

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