# Introduction to Coordinate Geometry: Exercise (1.1) - Solution

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.

Show/Hide Solution

2.           Find the missing coordinates in the following table if $M$ is the midpoint of points $P$ and $Q$. $$\begin{array}{|c|c|c|} \hline P & Q & M \\ \hline (2,6) & & (3,3) \\ \hline (3,2) & (-3,-1) & \\ \hline & (0,-1) & (-3,2) \\ \hline (1,5) & & (2.5,3.5) \\ \hline \end{array}$$

Show/Hide Solution

(i) Let the coordinates of the point $Q$ be $(a, b)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$(3,3) = \left(\displaystyle\frac{a+2}{2},\displaystyle \frac{b+6}{2}\right)$

$\therefore\ \displaystyle \frac{a+2}{2} = 3\ \text{and}\ \displaystyle \frac{b+6}{2} = 3$

$\therefore\ a=4\ \text{and}\ b = 0$

$\therefore\ Q=(4,0)$

(ii) Let the coordinates of the point $M$ be $(x, y)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

\begin{aligned} (x,y) &= \left(\frac{3-3}{2},\displaystyle \frac{2-1}{2}\right)\\\\ \therefore\ (x,y) &=\left(0,\frac{1}{2}\right) \end{aligned}

(iii) Let the coordinates of the point $P$ be $(x, y)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$(-3,2) = \left(\displaystyle\frac{x+0}{2},\displaystyle \frac{y-1}{2}\right)$

$\therefore\ \displaystyle \frac{x}{2} = -3\ \text{and}\ \displaystyle \frac{y-1}{2} = 2$

$\therefore\ x=-6\ \text{and}\ y = 5$

$\therefore\ P=(-6,5)$

(iv) Let the coordinates of the point $Q$ be $(x, y)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$(2.5,3.5) = \left(\displaystyle\frac{x+1}{2},\displaystyle \frac{y+5}{2}\right)$

$\therefore\ \displaystyle \frac{x+1}{2} = 2.5\ \text{and}\ \displaystyle \frac{y+5}{2} = 3.5$

$\therefore\ x=4\ \text{and}\ y = 2$

$\therefore\ P=(4,2)$

3.           Find the coordinates of the midpoint and the length of the line segment joining these pairs of points. $$\begin{array}{l} \text{(a)}\ (0,0)\ \text{and}\ (4, -4) \\ \text{(b)}\ (1, 5)\ \text{and}\ (3,1) \\ \text{(c)}\ (-3, -3)\ \text{and}\ (0,0) \\ \text{(d)}\ (-1,3)\ \text{and}\ (5, 1) \\ \text{(e)}\ (-1,6)\ \text{and}\ (2, -2) \\ \text{(f)}\ (-3, -4)\ \text{and}\ (3, -1) \end{array}$$

Show/Hide Solution

\begin{aligned} \text{(a)}\ & \text{The midpoint between } (0,0)\ \text{and }(4,-4)\\\\ &=\left( \frac{{0+4}}{2},\displaystyle \frac{{0-4}}{2}} \right)\\\\ &=\left( {2,-2} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(4-0)^2+(-4-0)^2}\\\\ &= 4\sqrt{2} \end{aligned

\begin{aligned} \text{(b)}\ & \text{The midpoint between }(1,5)\ \text{and }(3,1)\\\\ &=\left( \frac{{1+3}}{2},\displaystyle \frac{{5+1}}{2}} \right)\\\\ &=\left( {2,3} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(3-1)^2+(1-5)^2}\\\\ &= 2\sqrt{5} \end{aligned

\begin{aligned} \text{(c)}\ & \text{The midpoint between }(-3,-3)\ \text{and }(0,0)\\\\ &=\left( \frac{{-3+0}}{2},\displaystyle \frac{{-3+0}}{2}} \right)\\\\ &=\left( {-\displaystyle \frac{{3}}{2},-\displaystyle \frac{{3}}{2}} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(0+3)^2+(0+3)^2}\\\\ &=\sqrt{18}\\\\ &= 3\sqrt{2} \end{aligned

\begin{aligned} \text{(d)}\ & \text{The midpoint between }(-1,3)\ \text{and }(5,1)\\\\ &=\left( \frac{{-1+5}}{2},\displaystyle \frac{{3+1}}{2}} \right)\\\\ &=\left( {2,2} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(5+1)^2+(1-3)^2}\\\\ &=\sqrt{40}\\\\ &= 2\sqrt{10} \end{aligned

\begin{aligned} \text{(e)}\ & \text{The midpoint between }(-1,6)\ \text{and }(2,-2)\\\\ &=\left( \frac{{-1+2}}{2},\displaystyle \frac{{6-2}}{2}} \right)\\\\ &=\left(\displaystyle \displaystyle \frac{1}{2},2 \right)\\\\ &\text{length of segment}\\\\ &=\sqrt{(2+1)^2+(-2-6)^2}\\\\ &=\sqrt{73} \end{aligned

\begin{aligned} \text{(f)}\ &\text{The midpoint between }(-3,-4)\ \text{and }(3,-1)\\\\ &=\left( \frac{{-3+3}}{2},\displaystyle \frac{{-4-1}}{2}} \right)\\\\ &=\left( {0,-\displaystyle \frac{{5}}{2}} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(3+3)^2+(-1+4)^2}\\\\ &=\sqrt{45}\\\\ &= 3\sqrt{5} \end{aligned

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 $y$.

Show/Hide Solution

Since $(1,0)$ is the midpoint of $A(-5,2)$ and $B(x,y)$, using midpoint formula,

$(1,0) = \left(\displaystyle \frac{-5+x}{2}, \displaystyle \frac{2+y}{2}\right)$

$\therefore\ \displaystyle \frac{-5+x}{2} = 1\ \text{and}\ \displaystyle \frac{2+y}{2} = 0$

$\therefore\ x=7\ \text{and}\ y = -2$

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 $H(-5, 4)$.

Show/Hide Solution

(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}

$\therefore\ AB + BC + CD + AD =7.0+6.1+5.0+6.1=24.2$

6.           A circle has centre $(2, 1$). Find the coordinates of the endpoint of a diameter if one endpoint is $(7, 1)$.

Show/Hide Solution

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 $\triangle KLM$.

Show/Hide Solution

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}