1. Draw a set of coordinate axes. Locate the points, $A(2,3)$, $B(2, -4)$ and $C(-4, 3)$. Label each point with its coordinates. Determine whether each of the line segments $AB, BC$ and $CA$ is horizontal or vertical.
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2. Find the missing coordinates in the following table if $M$ is the midpoint of points $P$ and $Q$. $$\begin{array}{|c|c|c|} \hline P & Q & M \\ \hline (2,6) & & (3,3) \\ \hline (3,2) & (-3,-1) & \\ \hline & (0,-1) & (-3,2) \\ \hline (1,5) & & (2.5,3.5) \\ \hline \end{array}$$
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3. Find the coordinates of the midpoint and the length of the line segment joining these pairs of points. $$\begin{array}{l} \text{(a)}\ (0,0)\ \text{and}\ (4, -4) \\ \text{(b)}\ (1, 5)\ \text{and}\ (3,1) \\ \text{(c)}\ (-3, -3)\ \text{and}\ (0,0) \\ \text{(d)}\ (-1,3)\ \text{and}\ (5, 1) \\ \text{(e)}\ (-1,6)\ \text{and}\ (2, -2) \\ \text{(f)}\ (-3, -4)\ \text{and}\ (3, -1) \end{array}$$
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4. If $(1,0)$ is the midpoint of the line passing through the points $A(-5, 2)$ and $B(x, y)$, find the value of $x$ and of $y$.
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5. Calculate the perimeter of given polygons correct to one decimal place.
(a) A triangle with vertices $P(-2, 3), Q(5, -4)$ and $R(1, 8)$.
(b) A parallelogram with vertices $A(-10, 1), B(6, -2), C(14, 4)$ and $D(-2, 7)$.
(c) A trapezium with vertices $E(-6, -2), F(1, -2), G(0, 4)$ and $H(-5, 4)$.
(a) A triangle with vertices $P(-2, 3), Q(5, -4)$ and $R(1, 8)$.
(b) A parallelogram with vertices $A(-10, 1), B(6, -2), C(14, 4)$ and $D(-2, 7)$.
(c) A trapezium with vertices $E(-6, -2), F(1, -2), G(0, 4)$ and $H(-5, 4)$.
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(a) $ P=(-2,3),Q=(5,-4),R=(1,8)$
Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$
$\begin{aligned} PQ&=\sqrt{(5+2)^2+(-4-3)^2}\\\\ &=\sqrt{98}\\\\ &=9.9\\\\ QR &=\sqrt{(1-5)^2+(8+4)^2}\\\\ &=\sqrt{160}\\\\ &=12.7\\\\ PR&=\sqrt{(1+2)^2+(8-3)^2}\\\\ &=\sqrt{34}\\\\ &=5.8 \end{aligned}$
$\therefore\ PQ+QR+PR=28.4$
$\therefore$ the perimeter of $\triangle PQR = 28.4$ units.
(b) $A=(-10, 1), B=(6, -2), C=(14, 4), D=(-2, 7)$
Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$
$\begin{aligned} AB &=\sqrt{(6+10)^2+(-2-1)^2}\\\\ &=\sqrt{265}\\\\ &=16.3\\\\ BC &=\sqrt{(14-6)^2+(4+2)^2}\\\\ &=\sqrt{100}\\\\ &=10.0 \end{aligned}$
Since $ABCD$ is a parallelogram, $CD=AB$ and $AD = BC$
$\therefore\ AB + BC + CD + AD =2(16.3) + 2(10)=52.6$
$\therefore$ the perimeter of triangle parallelogram $ABCD = 52.6$ units.
(c) $E=(-6, -2), F=(1, -2), G=(0, 4), H=(-5, 4)$
Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$
$\begin{aligned} EF &=\sqrt{(1+6)^2+(-2+2)^2}\\\\ &=\sqrt{7^2}\\\\ &=7.0\\\\ FG &=\sqrt{(0-1)^2+(4+2)^2}\\\\ &=\sqrt{37}\\\\ &=6.1\\\\ GH &=\sqrt{(-5-0)^2+(4-4)^2}\\\\ &=\sqrt{5^2}\\\\ &=5.00\\\\ EH &=\sqrt{(-5+6)^2+(4+2)^2}\\\\ &=\sqrt{37}\\\\ &=6.1\\\\ \end{aligned}$
$\therefore\ AB + BC + CD + AD =7.0+6.1+5.0+6.1=24.2$
Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$
$\begin{aligned} PQ&=\sqrt{(5+2)^2+(-4-3)^2}\\\\ &=\sqrt{98}\\\\ &=9.9\\\\ QR &=\sqrt{(1-5)^2+(8+4)^2}\\\\ &=\sqrt{160}\\\\ &=12.7\\\\ PR&=\sqrt{(1+2)^2+(8-3)^2}\\\\ &=\sqrt{34}\\\\ &=5.8 \end{aligned}$
$\therefore\ PQ+QR+PR=28.4$
$\therefore$ the perimeter of $\triangle PQR = 28.4$ units.
(b) $A=(-10, 1), B=(6, -2), C=(14, 4), D=(-2, 7)$
Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$
$\begin{aligned} AB &=\sqrt{(6+10)^2+(-2-1)^2}\\\\ &=\sqrt{265}\\\\ &=16.3\\\\ BC &=\sqrt{(14-6)^2+(4+2)^2}\\\\ &=\sqrt{100}\\\\ &=10.0 \end{aligned}$
Since $ABCD$ is a parallelogram, $CD=AB$ and $AD = BC$
$\therefore\ AB + BC + CD + AD =2(16.3) + 2(10)=52.6$
$\therefore$ the perimeter of triangle parallelogram $ABCD = 52.6$ units.
(c) $E=(-6, -2), F=(1, -2), G=(0, 4), H=(-5, 4)$
Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$
$\begin{aligned} EF &=\sqrt{(1+6)^2+(-2+2)^2}\\\\ &=\sqrt{7^2}\\\\ &=7.0\\\\ FG &=\sqrt{(0-1)^2+(4+2)^2}\\\\ &=\sqrt{37}\\\\ &=6.1\\\\ GH &=\sqrt{(-5-0)^2+(4-4)^2}\\\\ &=\sqrt{5^2}\\\\ &=5.00\\\\ EH &=\sqrt{(-5+6)^2+(4+2)^2}\\\\ &=\sqrt{37}\\\\ &=6.1\\\\ \end{aligned}$
$\therefore\ AB + BC + CD + AD =7.0+6.1+5.0+6.1=24.2$
6. A circle has centre $(2, 1$). Find the coordinates of the endpoint of a diameter if one endpoint is $(7, 1)$.
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Let the other endpoint be $(x, y)$.
Hence, $(2, 1)$ is the midpoint between $(x, y)$ and $(7,1)$
Using midpoint formula,
$(2, 1) = \left(\frac{x+7}{2}, \frac{y+1}{2}\right)$
$\therefore\ \frac{x+7}{2} = 2\ \text{and}\ \frac{y+1}{2} = 1$
$\therefore\ x = -3\ \text{and}\ y = 1$
Hence the other endpoint is $(-3, 1).$
7. $\triangle KLM$ has vertices $K(-5,18)$, $L(10,14)$ and $M(-5, -10)$.
(a) Find the length of each side.
(b) Find the perimeter of $\triangle KLM$.
(c) Find the area of $\triangle KLM$.
(a) Find the length of each side.
(b) Find the perimeter of $\triangle KLM$.
(c) Find the area of $\triangle KLM$.
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Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$
$\begin{aligned} KL &=\sqrt{(10+5)^2+(14-18)^2}\\\\ &=\sqrt{241}\\\\ &=15.5\\\\ LM &=\sqrt{(-5-10)^2+(-10-14)^2}\\\\ &=\sqrt{801}\\\\ &=3\sqrt{89}\\\\ &=28.3\\\\ KM &=\sqrt{(-5+5)^2+(-10-18)^2}\\\\ &=\sqrt{28^2}\\\\ &=28\\\\ \text{The}\ & \text{perimeter of}\ \triangle KLM \\\\ & = KL+LM+KM\\\\ & = 15.5+28.3+28\\\\ &= 71.8\ \text{units} \end{aligned}$
Since $K$ and $M$ have the same $x$-coordinate, $KM$ is a vertical line.
Draw $LN\bot KM$.
Hence the coordinates of $N$ is $(-5,14)$.
$\begin{aligned} \therefore\ LN &=\sqrt{(-5-10)^2+(14-14)^2}\\\\ &=\sqrt{15^2}\\\\ &=15\\\\ \text{The}\ & \text{area of}\ \triangle KLM \\\\ & = \frac{1}{2}\cdot KM\cdot LN\\\\ & = \frac{1}{2}\cdot 28\cdot 15\\\\ & = 210\ \text{sq-units} \end{aligned}$
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