Long Questions (Question 5)

Question 5:
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)
$\begin{array}{l}2y=5x-21\\ y=\frac{5}{2}x-\frac{21}{2}\\ m_{AD}=\frac{5}{2}\\ m_{AB}\times m_{AD}=-1\\ m_{AB}\times \frac{5}{2}=-1\\ m_{AB}=-\frac{2}{5}\\ \\ \text{Equation of }AB\\ y-y_1=m_{AB}\left(x-x_1\right)\\ y+1=-\frac{2}{5}\left(x+2\right)\\ 5y+5=-2x-4\\ 5y=-2x-9\end{array}$

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

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