**Question 6:**The point

*M*(

*x*, 4), is the midpoint of the line joining straight line

*Q*(-2, -3) and

*R*(14,

*y*).

The value of

*x*and

*y*are

*Solution*:$\begin{array}{l}x=\frac{-2+14}{2}\\ x=\frac{12}{2}\\ x=6\\ \\ 4=\frac{-3+y}{2}\\ 8=-3+y\\ y=11\end{array}$

**Question 7:**In diagram below,

*PQR*is a right-angled triangle. The sides

*QR*and

*PQ*are parallel to the

*y*-axis and the

*x*-axis respectively. The length of

*QR*= 6 units.

Given that

*M*is the midpoint of

*PR*, then the coordinates of

*M*are

*Solution*:*x*-coordinate of

*R*= 3

*y*-coordinate of

*R*= 1 + 6 = 7

*R*= (3, 7)

$\begin{array}{l}P\left(1,1\right),R\left(3,7\right)\\ \text{Coordinatesof}M\\ =\left(\frac{1+3}{2},\frac{1+7}{2}\right)\\ =\left(2,4\right)\end{array}$

**Question 8:**Given points

*P*(–2, 8) and

*Q*(10, 8), find the length of

*PQ*.

*Solution*:$\begin{array}{l}\text{Lengthof}PQ\\ =\sqrt{{\left[10-\left(-2\right)\right]}^{2}+{\left(8-8\right)}^{2}}\\ =\sqrt{{\left(14\right)}^{2}+0}\\ =14\text{units}\end{array}$

**Question 9:**

In diagram below,

*ABC*is an isosceles triangle.

Find

(a) the value of

*k*,

(b) the length of

*BC*.

*Solution*:$\begin{array}{l}\left(\text{a}\right)\\ \text{Foranisoscelestriangle,}\\ y-\text{coordinateof}C\text{isthemidpointofstraightline}AB.\\ \frac{2+k}{2}=-3\\ 2+k=-6\\ \text{}k=-8\\ \\ \left(\text{b}\right)\\ B=\left(-2,-8\right)\\ BC=\sqrt{{\left[10-\left(-2\right)\right]}^{2}+{\left[-3-\left(-8\right)\right]}^{2}}\\ \text{}=\sqrt{{12}^{2}+{5}^{2}}\\ \text{}=13\text{units}\end{array}$

**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*:$\begin{array}{l}\text{Allsidesofrhombushavethesamelength,}\\ \text{thereforelengthofeachside}=\frac{40}{4}=10\text{units}\\ PQ=10\\ {\left(9-{x}_{1}\right)}^{2}+{\left(7-\left(-1\right)\right)}^{2}={10}^{2}\\ 81-18{x}_{1}+{x}_{1}{}^{2}+64=100\\ {x}_{1}{}^{2}-18{x}_{1}+45=0\\ \left({x}_{1}-3\right)\left({x}_{1}-15\right)=0\\ {x}_{1}=3,15\\ {x}_{1}=3\\ Q=\left(3,-1\right),R=\left({x}_{2},-1\right)\\ \\ QR=10\\ {\left({x}_{2}-3\right)}^{2}+{\left[-1-\left(-1\right)\right]}^{2}={10}^{2}\\ {x}_{2}{}^{2}-6{x}_{2}+9+0=100\\ {x}_{2}{}^{2}-6{x}_{2}-91=0\\ \left({x}_{2}+7\right)\left({x}_{2}-13\right)=0\\ {x}_{2}=-7,13\\ {x}_{2}=13\\ \\ \therefore R=\left(13,-1\right)\end{array}$