5.2.4 Coordinates, PT3 Focus Practice

Question 9:
In diagram below, ABC is an isosceles triangle.

Find
(a) the value of k,
(b) the length of BC.

Solution:
$\begin{array}{l}\left(\text{a}\right)\\ \text{For an isosceles triangle, }\\ y-\text{coordinate of }C\text{ is the midpoint of straight line }AB.\\ \frac{2+k}{2}=-3\\ 2+k=-6\\ k=-8\\ \\ \left(\text{b}\right)\\ B=\left(-2,-8\right)\\ BC=\sqrt{{\left[10-\left(-2\right)\right]}^2+{\left[-3-\left(-8\right)\right]}^2}\\ =\sqrt{{12}^2+5^2}\\ =13\text{ units}\end{array}$

Question 10:
Diagram below shows a rhombus PQRS drawn on a Cartesian plane. PS is parallel to x-axis.

Given the perimeter of PQRS is 40 units, find the coordinates of point R.

Solution:
$\begin{array}{l}\text{All sides of rhombus have the same length,}\\ \text{therefore length of each side}=\frac{40}{4}=10\text{ units}\\ PQ=10\\ {\left(9-x_1\right)}^2+{\left(7-\left(-1\right)\right)}^2={10}^2\\ 81-18x_1+x_1{}^2+64=100\\ x_1{}^2-18x_1+45=0\\ \left(x_1-3\right)\left(x_1-15\right)=0\\ x_1=3,15\\ x_1=3\\ Q=\left(3,-1\right),R=\left(x_2,-1\right)\\ \\ QR=10\\ {\left(x_2-3\right)}^2+{\left[-1-\left(-1\right)\right]}^2={10}^2\\ x_2{}^2-6x_2+9+0=100\\ x_2{}^2-6x_2-91=0\\ \left(x_2+7\right)\left(x_2-13\right)=0\\ x_2=-7,13\\ x_2=13\\ \\ \therefore R=\left(13,-1\right)\end{array}$