Space, shape and measurement: Use the Cartesian co-ordinate system to derive and apply equations

# Unit 2: Distance, gradient and midpoints

Dylan Busa

### Unit outcomes: Unit 2: Distance, gradient and midpoints

By the end of this unit you will be able to:

• Calculate the distance between two points on the Cartesian plane.
• Calculate the gradient between two points on a straight line.
• Work with the gradients of parallel and perpendicular lines.
• Calculate the midpoint between two points on the Cartesian plane.

## What you should know

Before you start this unit, make sure you can

## Introduction

In Unit 1 of this Subject outcome we plotted points on the Cartesian plane to draw line segments and polygons. One type of polygon that we could draw is a square. But how can we prove that the shape we have drawn is really a square?

To prove that a polygon is a square, we need to prove that all the sides are the same length AND that the interior angles are all equal to $\scriptsize {{90}^{{}^\circ }}$. We could take some measurements with a ruler and protractor, but this is not always accurate or practical.

We need better methods to analyse the lines and shapes we draw on the Cartesian plane. Let’s start with the lengths of lines.

## The distance between two points

Being able to calculate the distance between any two points on the Cartesian plane is very useful and surprisingly easy as you will discover in Activity 2.1.

### Activity 2.1: Measure distance on the Cartesian plane

Time required: 15 minutes

What you need:

• a piece of paper (graph paper if possible)
• a pen or pencil

What to do:

1. Draw your own Cartesian plane on a piece of paper. Use graph paper if you have some. Plot the following points: $\scriptsize \text{A}(2,3)$, $\scriptsize \text{B}(2,0)$ and $\scriptsize \text{C}(-2,0)$. Now join the points $\scriptsize \text{A, B}$ and $\scriptsize \text{C}$. What type of shape have you created?
2. How do you know that the angle at $\scriptsize \text{B}$ is $\scriptsize {{90}^{{}^\circ }}$?
3. What is the length of $\scriptsize \text{AB}$? How do you know?
4. What is the length of $\scriptsize \text{BC}$? How do you know?
5. How can we find the length of $\scriptsize \text{AC}$ in this right-angled triangle?
6. Use the theorem of Pythagoras to find the length of $\scriptsize \text{AC}$.
7. If point $\scriptsize \text{A}$ had coordinates $\scriptsize ({{x}_{1}},{{y}_{1}})$, point $\scriptsize \text{B}$ had coordinates $\scriptsize ({{x}_{1}},{{y}_{2}})$, point and $\scriptsize \text{C}$ had coordinates $\scriptsize ({{x}_{2}},{{y}_{2}})$, write expressions for the lengths of $\scriptsize \text{AB}$ and $\scriptsize \text{BC}$.
8. Now write an expression to find the length of $\scriptsize \text{AC}$.

What you found:

1. $\scriptsize \text{ABC}$ is a triangle.
2. We know that the axes are perpendicular to each other. The line $\scriptsize \text{AB}$ is parallel to the y-axis because both points have the same x-coordinate. The line $\scriptsize \text{BC}$ is parallel to the x-axis because it lies on the x-axis. Therefore, we know that $\scriptsize \text{AB}$ is perpendicular to $\scriptsize \text{BC}$.
3. Since $\scriptsize \text{A}$ and $\scriptsize \text{B}$ lie on the same vertical line, the length of $\scriptsize \text{AB}$ is the difference in the y-coordinates. Therefore, the length of $\scriptsize \text{AB}$ is $\scriptsize 3-0=3$.
4. Since $\scriptsize \text{B}$ and $\scriptsize \text{C}$ lie on the same horizontal line, the length of $\scriptsize \text{BC}$ is the difference in the x-coordinates. Therefore, the length of $\scriptsize \text{BC}$ is $\scriptsize 2-(-2)=2+2=4$.
5. In $\scriptsize \Delta \text{ABC}$, $\scriptsize \text{AC}$ lies opposite the $\scriptsize {{90}^{{}^\circ }}$ angle and is therefore the hypotenuse. We can use the theorem of Pythagoras to calculate the length of $\scriptsize \text{AC}$.
6. In $\scriptsize \Delta \text{ABC}$:
\scriptsize \begin{align*}\text{A}{{\text{C}}^{2}} & =\text{A}{{\text{B}}^{2}}+\text{B}{{\text{C}}^{2}}\text{ Pythagoras}\\\therefore \text{A}{{\text{C}}^{2}} & ={{3}^{2}}+{{4}^{2}}\\\therefore \text{A}{{\text{C}}^{2}} & =25\\\therefore \text{AC} & =5\end{align*}
7. The length of $\scriptsize \text{AB}$ is $\scriptsize {{y}_{1}}-{{y}_{2}}$. The length of $\scriptsize \text{BC}$ is $\scriptsize {{x}_{1}}-{{x}_{2}}$.
8. In $\scriptsize \Delta \text{ABC}$:
\scriptsize \begin{align*}\text{A}{{\text{C}}^{2}} & =\text{A}{{\text{B}}^{2}}+\text{B}{{\text{C}}^{2}}\\\therefore \text{A}{{\text{C}}^{2}} & ={{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{x}_{1}}-{{x}_{2}})}^{2}}\\\therefore \text{AC} & =\sqrt{{{{{({{y}_{1}}-{{y}_{2}})}}^{2}}+{{{({{x}_{1}}-{{x}_{2}})}}^{2}}}}=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\end{align*}

We have just seen that to calculate the distance between any two points $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize ({{x}_{2}},{{y}_{2}})$ on the Cartesian plane we can use the formula $\scriptsize d=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}$. This is called the distance formula.

Distance formula: $\scriptsize d=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}$.

### Note

Because $\scriptsize {{({{x}_{1}}-{{x}_{2}})}^{2}}={{({{x}_{2}}-{{x}_{1}})}^{2}}$ it does not matter which point we assign as point $\scriptsize 1$ and which we assign as point $\scriptsize 2$ in $\scriptsize d=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}$

### Example 2.1

If $\scriptsize \text{H}(-3,6)$, $\scriptsize \text{I}(8,4)$ and $\scriptsize \text{J}(-3,-5)$, find the distance between $\scriptsize \text{H}$ and $\scriptsize \text{I}$ and between $\scriptsize \text{I}$ and $\scriptsize \text{J}$.

Solution

Distance between $\scriptsize \text{H}$ and $\scriptsize \text{I}$:
The first step is to assign each point to be either point $\scriptsize 1$ or point $\scriptsize 2$. Let $\scriptsize \text{H}$ be $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{I}$ be $\scriptsize ({{x}_{2}},{{y}_{2}})$.

The distance formula is $\scriptsize d=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}$. Substitute the given coordinates into the formula.

\scriptsize \begin{align*}{{d}_{{\text{HI}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-3-8)}}^{2}}+{{{(6-4)}}^{2}}}}\\&=\sqrt{{{{{(-11)}}^{2}}+{{{(2)}}^{2}}}}\\&=\sqrt{{121+4}}\\&=\sqrt{{125}}\\&=11.2\quad \text{rounded of to 1 decimal place}\end{align*}

The distance between $\scriptsize \text{H}(-3,6)$ and $\scriptsize \text{I}(8,4)$ is $\scriptsize 11.2$ units.

Distance between $\scriptsize \text{I}$ and $\scriptsize \text{J}$:
The first step again is to assign each point to be either point $\scriptsize 1$ or point $\scriptsize 2$. Now let $\scriptsize \text{I}$ be $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{J}$ be $\scriptsize ({{x}_{2}},{{y}_{2}})$.

The distance formula is $\scriptsize d=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}$. Substitute the given coordinates into the formula.

\scriptsize \begin{align*}{{d}_{{\text{IJ}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(8-(-3))}}^{2}}+{{{(4-(-5))}}^{2}}}}\end{align*}

It is always a good idea to place negative values inside brackets (like we did with the coordinates of $\scriptsize \text{J}$) so that you don’t make any mistakes with the signs.

\scriptsize \begin{align*}{{d}_{{\text{IJ}}}} & =\sqrt{{{{{(8-(-3))}}^{2}}+{{{(4-(-5))}}^{2}}}}\\&=\sqrt{{{{{(8+3)}}^{2}}+{{{(4+5)}}^{2}}}}\\&=\sqrt{{{{{11}}^{2}}+{{9}^{2}}}}\\&=\sqrt{{121+81}}\\&=\sqrt{{202}}\\&=14.2\quad \text{rounded of to 1 decimal place}\end{align*}

### Example 2.2

If $\scriptsize \text{EF}=4\sqrt{5}$, $\scriptsize \text{E}(-2,3)$ and $\scriptsize \text{F}(x,-5)$, find the value of $\scriptsize x$.

Solution

It is a good idea to make a sketch of the given information to help you think through the problem.

We know that the point $\scriptsize \text{F}$ lies somewhere on the horizontal line that cuts the y-axis at $\scriptsize -5$ but there are two possible places where it could be ($\scriptsize {{\text{F}}_{1}}$ or $\scriptsize {{\text{F}}_{2}}$).

We need to assign each point to be either point $\scriptsize 1$ or point $\scriptsize 2$. Let $\scriptsize \text{E}$ be $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{F}$ be $\scriptsize ({{x}_{2}},{{y}_{2}})$.

The distance formula is $\scriptsize d=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}$. Substitute the given coordinates into the formula.

\scriptsize \begin{align*}{{d}_{{\text{EF}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-2-x)}}^{2}}+{{{(3-(-5))}}^{2}}}}\\&=\sqrt{{{{{(-2-x)}}^{2}}+{{{(8)}}^{2}}}}\\&=\sqrt{{{{{(-2-x)}}^{2}}+64}}\end{align*}

But we know that $\scriptsize {{d}_{{\text{EF}}}}=4\sqrt{5}$.

\scriptsize \begin{align*}4\sqrt{5} & =\sqrt{{{{{(-2-x)}}^{2}}+64}}\\\therefore {{\left( {4\sqrt{5}} \right)}^{2}} & ={{(-2-x)}^{2}}+64\quad \text{Expand }{{(-2-x)}^{2}}\\\therefore 80 & =4+4x+{{x}^{2}}+64\quad \text{Solve the quadratic equation}\\\therefore {{x}^{2}}+4x-12 & =0\\\therefore (x+6)(x-2) & =0\\\therefore x=-6\text{ } & \text{or }x=2\end{align*}

As we expected, we get two possible answers.

### Note

Drawing a sketch of the situation helps you to make sense of the information given as well as to check the feasibility of your answers.

### Exercise 2.1

1. Calculate the length of $\scriptsize \text{AB}$. Round your answer to one decimal place.
2. Find the distance between $\scriptsize \text{C}(0,-9)$ and $\scriptsize \text{D}(2,3)$. Leave your answer in surd form.
3. Calculate $\scriptsize {{d}_{{\text{QT}}}}$ if $\scriptsize \text{Q}(x,y)$ and $\scriptsize \text{T}(x+4,y-7)$. Leave your answer in surd form.
4. If the distance between $\scriptsize \text{S}(0,-3)$ and $\scriptsize \text{F}(8,a)$ is 10 units, find the possible values of $\scriptsize a$.

The full solutions are at the end of the unit.

### Note

If you would like more practise using the distance formula to find the length of a line on the Cartesian plane, try the interactive activity called Practice the Distance Formula.

We learnt about gradient ($\scriptsize m$) in Subject outcome 2.1 Unit 1 on linear functions. Gradient is a measure of the steepness of a line. In Figure 1, $\scriptsize \text{OA}$ has a steeper gradient than $\scriptsize \text{OB}$. While $\scriptsize \text{OA}$ and $\scriptsize \text{OB}$ have positive gradients, $\scriptsize \text{OC}$ and $\scriptsize \text{OD}$ have negative gradients.

We define the gradient of a line as the ratio of the change in $\scriptsize y$ values between any two points on the line (the rise or the vertical movement up or down) to the change in $\scriptsize x$ values between those same two points (the run or the horizontal movement left or right). We can write this ratio as a fraction like this:

$\scriptsize m=\displaystyle \frac{{\text{change in }y}}{{\text{change in }x}}=\displaystyle \frac{{\text{rise}}}{{\text{run}}}$.

To work out the change in $\scriptsize y$ or rise between any two points on the Cartesian plane, we simply need to find the difference between their y-coordinates. We find the difference between their x-coordinates to work out the change in $\scriptsize x$ or run. Therefore, the gradient between any two points on the Cartesian plane can be expressed as follows:

$\scriptsize m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$ or $\scriptsize m=\displaystyle \frac{{{{y}_{1}}-{{y}_{2}}}}{{{{x}_{1}}-{{x}_{2}}}}$

We call this the gradient formula.

Gradient formula: $\scriptsize m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$ or $\scriptsize m=\displaystyle \frac{{{{y}_{1}}-{{y}_{2}}}}{{{{x}_{1}}-{{x}_{2}}}}$

### Note

The order of the points does not matter but you must be consistent.
$\scriptsize m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$ or $\scriptsize m=\displaystyle \frac{{{{y}_{1}}-{{y}_{2}}}}{{{{x}_{1}}-{{x}_{2}}}}$

$\scriptsize m\ne \displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{1}}-{{x}_{2}}}}$ and $\scriptsize m\ne \displaystyle \frac{{{{y}_{1}}-{{y}_{2}}}}{{{{x}_{2}}-{{x}_{1}}}}$

### Example 2.3

Find the gradient between $\scriptsize \text{W}(5,2)$ and $\scriptsize \text{V}(9,-3)$.

Solution

If you want to, you can draw a sketch of the given information. This can help you to see if your answer makes sense.

We can see that the line slopes from left to right so we expect to get a negative answer for the gradient.

Let $\scriptsize \text{W}(5,2)$ be $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{V}(9,-3)$ be $\scriptsize ({{x}_{2}},{{y}_{2}})$.
\scriptsize \begin{align*}{{m}_{{\text{WV}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{-3-2}}{{9-5}}\\ & =\displaystyle \frac{{-5}}{4}\\ & =-\displaystyle \frac{5}{4}\end{align*}
Therefore $\scriptsize {{m}_{{\text{WV}}}}=-\displaystyle \frac{5}{4}$. We get a negative answer as expected.

### Note

If you would like more practise finding the gradient between two points, visit the interactive simulation called Calculating gradient.

Lines that run horizontally (parallel with the x-axis) have a gradient of zero. Since the y-values are the same everywhere on the line, the change in $\scriptsize y$ (the rise) of the line is zero.

$\scriptsize m=\displaystyle \frac{{\text{change in }y}}{{\text{change in }x}}=\displaystyle \frac{{\text{rise}}}{{\text{run}}}=\displaystyle \frac{0}{{\text{run}}}=0$. In this case $\scriptsize {{y}_{2}}-{{y}_{1}}=0$.

Lines that run vertically (parallel with the y-axis) have an undefined gradient. Here the change in $\scriptsize x$ (the run) is zero since the x-values are the same everywhere on the line but we are never allowed to divide by zero.

$\scriptsize m=\displaystyle \frac{{\text{change in }y}}{{\text{change in }x}}=\displaystyle \frac{{\text{rise}}}{{\text{run}}}=\displaystyle \frac{{\text{rise}}}{0}$ which is undefined.

$\scriptsize {{m}_{{\text{horizontal line}}}}=0$

$\scriptsize {{m}_{{\text{vertical line}}}}$ is undefined

### Example 2.4

$\scriptsize {{m}_{{\text{MN}}}}=\displaystyle \frac{5}{3}$ and $\scriptsize \text{M}(-2,4)$. Find $\scriptsize a$ if $\scriptsize \text{N}\left( {\displaystyle \frac{3}{2},a} \right)$.

Solution

We know that $\scriptsize m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}=\displaystyle \frac{5}{3}$. Let $\scriptsize \text{M}(-2,4)$ be $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{N}\left( {\displaystyle \frac{3}{2},a} \right)$ be $\scriptsize ({{x}_{2}},{{y}_{2}})$.
\scriptsize \begin{align*}m & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}=\displaystyle \frac{5}{3}\\\therefore \displaystyle \frac{5}{3} & =\displaystyle \frac{{a-4}}{{\displaystyle \frac{3}{2}-(-2)}}\\ & =\displaystyle \frac{{a-4}}{{\displaystyle \frac{{3+4}}{2}}}\quad \text{Add the fractions in the denominator}\\ & =\left(a-4\right)\times \displaystyle \frac{2}{7}\quad \text{Simplify by muliplying by the inverse of the denominator}\\\therefore \displaystyle \frac{5}{3} & =\displaystyle \frac{{2a-8}}{7}\\\therefore 5\times 7 & =3(2a-8)\\\therefore 35 & =6a-24\\\therefore 6a & =59\\\therefore a & =\displaystyle \frac{{59}}{6}=9\displaystyle \frac{5}{6}\end{align*}

### Exercise 2.2

1. Find the gradient of $\scriptsize \text{GH}$ if:
1. $\scriptsize \text{G}(3,-5)$ and $\scriptsize \text{H}(-2,7)$.
2. $\scriptsize \text{G}\left( {\displaystyle \frac{3}{2},0} \right)$ and $\scriptsize \text{H}\left( {3,-\displaystyle \frac{3}{4}} \right)$.
3. $\scriptsize \text{G}(x-3,y)$ and $\scriptsize \text{H}(x,y-4)$.
2. If the gradient of $\scriptsize {{m}_{{\text{ST}}}}=\displaystyle \frac{2}{3}$, find $\scriptsize q$ if:
1. $\scriptsize \text{S}(8,q)$ and $\scriptsize \text{T}(16,2)$.
2. $\scriptsize \text{S}(3,2q)$ and $\scriptsize \text{T}(9,14)$.

The full solutions are at the end of the unit.

## Parallel and perpendicular lines

We know that parallel lines are always the same distance apart. This means that on the Cartesian plane, parallel lines have the same gradient.

In Figure 2, we have two parallel lines. If we measure the gradient of one line between points $\scriptsize \text{A}$ and $\scriptsize \text{B}$, we can see that the gradient is $\scriptsize \displaystyle \frac{2}{4}=\displaystyle \frac{1}{2}$. If we measure the gradient of the other line between points $\scriptsize \text{C}$ and $\scriptsize \text{D}$, we can see that that gradient is $\scriptsize \displaystyle \frac{4}{8}=\displaystyle \frac{1}{2}$.

The gradients of these two lines are the same. This makes sense. If parallel lines never meet, they must have exactly the same slope, in other words, the same gradient.

We also know that perpendicular lines are at $\scriptsize {{90}^{{}^\circ }}$ to each other. That is what perpendicular means. But what is the relationship between the gradients of perpendicular lines? We will discover this in Activity 2.2.

### Activity 2.2: The gradients of perpendicular lines

Time required: 10 minutes

What you need:

• a piece of paper (graph paper if possible)
• a pen or pencil

What to do:

1. On your own Cartesian plane, draw $\scriptsize \text{ABC}$ such that $\scriptsize \text{A}(-3,4)$, $\scriptsize \text{B}(-4,1)$ and $\scriptsize \text{C}(3,2)$.
2. Calculate the lengths of $\scriptsize \text{AB}$, $\scriptsize \text{AC}$ and $\scriptsize \text{BC}$.
3. Use the lengths of the sides and the theorem of Pythagoras to prove that the angle at $\scriptsize \text{A}$ is $\scriptsize {{90}^{{}^\circ }}$.
4. Calculate the gradients of $\scriptsize \text{AB}$ and $\scriptsize \text{AC}$. What do you notice about the gradients of $\scriptsize \text{AB}$ and $\scriptsize \text{AC}$?

What you found:

1. Here is a sketch of $\scriptsize \Delta \text{ABC}$.
2. .
\scriptsize \begin{align*}{{d}_{{\text{AB}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-3-(-4))}}^{2}}+{{{(4-1)}}^{2}}}}\\&=\sqrt{{{{{(1)}}^{2}}+{{{(3)}}^{2}}}}\\&=\sqrt{{10}}\end{align*}\scriptsize \begin{align*}{{d}_{{\text{AC}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-3-3)}}^{2}}+{{{(4-2)}}^{2}}}}\\&=\sqrt{{{{{(-6)}}^{2}}+{{{(2)}}^{2}}}}\\&=\sqrt{{40}}\end{align*}\scriptsize \begin{align*}{{d}_{{\text{BC}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-4-3)}}^{2}}+{{{(1-2)}}^{2}}}}\\&=\sqrt{{{{{(-7)}}^{2}}+{{{(-1)}}^{2}}}}\\&=\sqrt{{50}}\end{align*}
3. If the angle at $\scriptsize \hat{\text{A}}={{90}^{{}^\circ }}$ then $\scriptsize \text{A}{{\text{B}}^{2}}+\text{A}{{\text{C}}^{2}}=\text{B}{{\text{C}}^{2}}$.
\scriptsize \begin{align*}\text{A}{{\text{B}}^{2}}+\text{A}{{\text{C}}^{2}} & =10+40\\ & =50\\ & =\text{B}{{\text{C}}^{2}}\end{align*}
Therefore $\scriptsize \hat{\text{A}}={{90}^{{}^\circ }}$.
4. .
\scriptsize \begin{align*}{{m}_{{\text{AB}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{4-1}}{{-3-(-4)}}\\ & =\displaystyle \frac{3}{1}\\ & =3\end{align*}
\scriptsize \begin{align*}{{m}_{{\text{AC}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{4-2}}{{-3-3}}\\ & =\displaystyle \frac{2}{{-6}}\\ & =-\displaystyle \frac{1}{3}\end{align*}
We can see that $\scriptsize {{m}_{{\text{AC}}}}$ is the negative inverse of $\scriptsize {{m}_{{\text{AB}}}}$. This means that $\scriptsize {{m}_{{\text{AB}}}}\times {{m}_{{\text{AC}}}}=-1$.
5. The product of the gradients of perpendicular lines is $\scriptsize -1$.

Activity 2.2 showed us that if two lines are perpendicular to each other then, when you multiply their gradients together, the answer is $\scriptsize -1$.

If $\scriptsize \text{AB}\parallel \text{CD}$ then $\scriptsize {{m}_{{\text{AB}}}}={{m}_{{\text{CD}}}}$

If $\scriptsize \text{AB}\bot \text{CD}$ then $\scriptsize {{m}_{{\text{AB}}}}\times {{m}_{{\text{CD}}}}=-1$

### Example 2.5

Prove that line $\scriptsize \text{AB}$ with $\scriptsize \text{A}(0,2)$ and $\scriptsize \text{B}(2,6)$ is parallel to line $\scriptsize \text{CD}$ with equation $\scriptsize 2x=y-7$.

Solution

First we need to work out the gradient of $\scriptsize \text{AB}$.
\scriptsize \begin{align*}{{m}_{{\text{AB}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{6-2}}{{2-0}}\\ & =\displaystyle \frac{4}{2}\\&=2\end{align*}

Next we need to work out the gradient of the line $\scriptsize \text{CD}$ based on the equation of the function given.

\scriptsize \begin{align*}2x & =y-7\\\therefore y & =2x+7\end{align*}

Therefore the gradient of $\scriptsize \text{CD}$ is $\scriptsize 2$. So $\scriptsize {{m}_{{\text{AB}}}}={{m}_{{\text{CD}}}}$. Therefore the lines are parallel.

### Example 2.6

Prove that $\scriptsize \text{Q}(2,-7)$ lies on the line $\scriptsize \text{PR}$ where $\scriptsize \text{P}(1,-3)$ and $\scriptsize \text{R}\left( {2\displaystyle \frac{1}{2},-9} \right)$.

Solution

If $\scriptsize \text{Q}$ lies on $\scriptsize \text{PR}$, then $\scriptsize {{m}_{{\text{PQ}}}}={{m}_{{\text{QR}}}}={{m}_{{\text{PR}}}}$.

\scriptsize \begin{align*}{{m}_{{\text{PQ}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\&=\displaystyle \frac{{-3-(-7)}}{{1-2}}\\&=\displaystyle \frac{{-3+7}}{{-1}}\\&=-4\end{align*}

\scriptsize \begin{align*}{{m}_{{\text{QR}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\&=\displaystyle \frac{{-7-(-9)}}{{2-2\displaystyle \frac{1}{2}}}\\&=\displaystyle \frac{{-7+9}}{{-\displaystyle \frac{1}{2}}}\\&=\displaystyle \frac{2}{1}\times \left( {-\displaystyle \frac{2}{1}} \right)\\&=-4\end{align*}

\scriptsize \begin{align*}{{m}_{{\text{PR}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\&=\displaystyle \frac{{-3-(-9)}}{{1-2\displaystyle \frac{1}{2}}}\\&=\displaystyle \frac{{-3+9}}{{-\displaystyle \frac{3}{2}}}\\&=\displaystyle \frac{6}{1}\times \left( {-\displaystyle \frac{2}{3}} \right)\\&=-\displaystyle \frac{{12}}{3}=-4\end{align*}

Therefore $\scriptsize {{m}_{{\text{PQ}}}}={{m}_{{\text{QR}}}}={{m}_{{\text{PR}}}}$ and so $\scriptsize \text{Q}$ lies on $\scriptsize \text{PR}$. We say that the points $\scriptsize \text{P}$, $\scriptsize \text{Q}$, and $\scriptsize \text{R}$ are collinear because they all lie on the same straight line.

### Note

Points are collinear if they lie on the same straight line.

### Example 2.7

Line $\scriptsize \text{KL}$ is perpendicular to line $\scriptsize \text{BC}$. Find $\scriptsize a$ if $\scriptsize \text{B}(2,-3)$, $\scriptsize \text{C}(-2,6)$, $\scriptsize \text{K}(4,3)$ and $\scriptsize \text{L}(7,a)$.

Solution

We are told that line $\scriptsize \text{KL}$ is perpendicular to line $\scriptsize \text{BC}$. Therefore we know that $\scriptsize {{m}_{{\text{KL}}}}\times {{m}_{{\text{BC}}}}=-1$.

First we need to calculate $\scriptsize {{m}_{{\text{BC}}}}$.

\scriptsize \begin{align*}{{m}_{{\text{BC}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\&=\displaystyle \frac{{-3-6}}{{2-(-2)}}\\&=\displaystyle \frac{{-9}}{4}\end{align*}

Therefore, we know that $\scriptsize {{m}_{{\text{KL}}}}=\displaystyle \frac{4}{9}$. We can use this to solve for $\scriptsize a$.

\scriptsize \begin{align*}{{m}_{{\text{BC}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\\therefore \displaystyle \frac{4}{9} & =\displaystyle \frac{{3-a}}{{4-7}}\\\therefore \displaystyle \frac{4}{9} & =\displaystyle \frac{{3-a}}{{-3}}\\\therefore -12 & =9(3-a)\\\therefore -12 & =27-9a\\\therefore 9a & =27+12=39\\\therefore a & =\displaystyle \frac{{39}}{9}=4\displaystyle \frac{3}{9}=4\displaystyle \frac{1}{3}\end{align*}

### Exercise 2.3

1. Determine whether $\scriptsize \text{AB}$ and $\scriptsize \text{CD}$ are parallel, perpendicular or neither.
1. $\scriptsize \text{A}(-1,-1)$, $\scriptsize \text{B}(0,-4)$, $\scriptsize \text{C}(3,-4)$, $\scriptsize \text{D}(5,2)$
2. $\scriptsize \text{A}(-1,3)$, $\scriptsize \text{B}(-2,2)$, $\scriptsize \text{C}(3,-4)$, $\scriptsize \text{D}(5,2)$
2. Do $\scriptsize \text{K}(-6,2)$, $\scriptsize \text{L}(-3,1)$ and $\scriptsize \text{M}(1,-1)$ lie on the same straight line?
3. $\scriptsize \text{JK}$ is perpendicular to the line given by $\scriptsize 4x-3y=9$. Find $\scriptsize q$ if $\scriptsize \text{J}(-3,1)$ and $\scriptsize \text{K}(q,-3)$.

The full solutions are at the end of the unit.

## The midpoint of a line

Sometimes it is necessary to find the middle or midpoint of a line segment on the Cartesian plane. Work through Activity 2.3 to find out how to do this.

### Activity 2.3: Find the midpoint of a line segment

Time required: 10 minutes

What you need:

• a piece of paper (graph paper if possible)
• a pen or pencil

What to do:

1. On your own Cartesina plane plot the points $\scriptsize \text{A}(-6,-2)$ and $\scriptsize \text{B}(5,7)$ and draw the line $\scriptsize \text{AB}$. Fold your piece of paper so that points $\scriptsize \text{A}$ and $\scriptsize \text{B}$ are exactly on top of each other. What are the coordinates of point $\scriptsize \text{C}$, the point where the fold line cuts line $\scriptsize \text{AB}$?
2. What can you say about the position of $\scriptsize \text{C}$ in relation to line $\scriptsize \text{AB}$?
3. Write the coordinates of point $\scriptsize \text{C}$ in terms of the coordinates of points $\scriptsize \text{A}$ and $\scriptsize \text{B}$.
4. Write a general expression for finding the coordinates of the midpoint of a line segment with $\scriptsize \text{A}({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{B}({{x}_{2}},{{y}_{2}})$.

What you found:

1. The coordinates of $\scriptsize \text{C}$ are $\scriptsize (-1,2)$. The dotted line in the following graph shows its location on the Cartesian plane.
2. Point $\scriptsize \text{C}$ is the middle of the line segment $\scriptsize \text{AB}$.
3. The x-coordinate of $\scriptsize \text{C}$ is the sum of the x-coordinates of $\scriptsize \text{A}$ and $\scriptsize \text{B}$ divided by $\scriptsize 2$: $\scriptsize \displaystyle \frac{{-6+4}}{2}=-1$
The y-coordinate of $\scriptsize \text{C}$ is the sum of the y-coordinates of $\scriptsize \text{A}$ and $\scriptsize \text{B}$ divided by $\scriptsize 2$: $\scriptsize \displaystyle \frac{{-2+6}}{2}=2$
4. $\scriptsize \text{C}\left( {\displaystyle \frac{{{{x}_{\text{A}}}+{{x}_{\text{B}}}}}{2},\displaystyle \frac{{{{y}_{\text{A}}}+{{y}_{\text{B}}}}}{2}} \right)$
5. The midpoint of $\scriptsize \text{AB}$ is given by $\scriptsize \left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)$

The midpoint of a line segemnt is $\scriptsize \text{M}(x,y)=\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)$. It does not matter which point is made point $\scriptsize 1$ or point $\scriptsize 2$.

### Example 2.8

Find the midpoint $\scriptsize \text{M}(x,y)$ of $\scriptsize \text{ST}$ where $\scriptsize \text{S}(4,0)$ and $\scriptsize \text{T}(-5,6)$. In which quadrant does the midpoint lie?

Solution

$\scriptsize \text{M}(x,y)=\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)$. Let $\scriptsize \text{S}$ be point $\scriptsize 1$.
\scriptsize \begin{align*}\text{M}(x,y) & =\left( {\displaystyle \frac{{4-5}}{2},\displaystyle \frac{{0+6}}{2}} \right)\\ & =\left( {-\displaystyle \frac{1}{2},3} \right)\end{align*}
The midpoint lies in quadrant II.

### Example

$\scriptsize \text{CD}$ has a midpoint $\scriptsize \text{M}(1,-3)$. Find the point $\scriptsize \text{C}$ if $\scriptsize \text{D}(6,5)$.

Solution

In this case we are given the midpoint $\scriptsize \text{M}(x,y)$ and we know that $\scriptsize \text{M}(x,y)=\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)$. We can calculate the x- and y-coordinates of $\scriptsize \text{C}$ seperately. Let $\scriptsize \text{D}(6,5)$ be point $\scriptsize 1$.

The x-coordinate of $\scriptsize \text{C}$:
\scriptsize \begin{align*}1 & =\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2}\\\therefore 1 & =\displaystyle \frac{{6+{{x}_{2}}}}{2}\\\therefore 2 & =6+{{x}_{2}}\\\therefore {{x}_{2}} & =-4\end{align*}

The y-coordinate of $\scriptsize \text{C}$:
\scriptsize \begin{align*}-3 & =\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}\\\therefore -3 & =\displaystyle \frac{{5+{{y}_{2}}}}{2}\\\therefore -6 & =6+{{y}_{2}}\\\therefore {{y}_{2}} & =-12\end{align*}

### Exercise 2.4

1. Find the midpoint of $\scriptsize \text{AB}$ where $\scriptsize \text{A}(-4,7)$ and $\scriptsize \text{B}(2,5)$.
2. Find the midpoint of $\scriptsize \text{GH}$ where $\scriptsize \text{G}(x+3,y-2)$ and $\scriptsize \text{H}(x-5,y-4)$.
3. The midpoint $\scriptsize \text{M}$ of $\scriptsize \text{PQ}$ is $\scriptsize (3,9)$. Find $\scriptsize \text{Q}$ if $\scriptsize \text{P}(1,1)$.

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to calculate the distance between two points on the Cartesian plane using the distance formula.
• How to calculate the gradient between any two points on a straight line.
• How to work with the gradients of parallel and perpendicular lines.
• That the gradients of parallel lines are the same.
• That the product of the gradients of perpendicular lines is equal to $\scriptsize -1$.
• How to calculate the midpoint between two points on the Cartesian plane.

# Unit 2: Assessment

#### Suggested time to complete: 45 minutes

1. Line $\scriptsize \text{AB}$ is shown below. Calculate its:
1. Length (correct to one decimal place)
3. Midpoint
2. If $\scriptsize \text{CD}=5$, find $\scriptsize a$ if $\scriptsize \text{D}\left( {-1,a} \right)$.
3. $\scriptsize \text{ABCD}$ is a parellogram where $\scriptsize \text{A}(5,3)$, $\scriptsize \text{B}(2,1)$ and $\scriptsize \text{C}(7,-3)$. Find $\scriptsize \text{D}$.
HINT: The diagonals of a parallelogram bisect each other (they cut each other in half). Start by drawing a diagram of the information you have been given.
4. The vertices of $\scriptsize \text{EFGH}$ are shown below.
1. Calculate the lengths of the sides of $\scriptsize \text{EFGH}$.
2. Are the opposite sides of $\scriptsize \text{EFGH}$ parallel?
3. Do the diagonals of $\scriptsize \text{EFGH}$ bisect each other?
4. What kind of quadrilateral is $\scriptsize \text{EFGH}$? Give reasons for your answer.
5. Question 5 adapted from Everything Maths Grade 10 Exercise 8-6 Question 37
$\scriptsize \text{A}(-2,4)$, $\scriptsize \text{B}(-4,-2)$and $\scriptsize \text{C}(4,0)$ are the vertices of $\scriptsize \Delta \text{ABC}$. $\scriptsize \text{D}$ and $\scriptsize \text{E}(1,2)$ are the midpoints of $\scriptsize \text{AB}$ and $\scriptsize \text{AC}$ respectively.
1. Find the gradient of $\scriptsize \text{BC}$.
2. Show that $\scriptsize \text{D}$ is the point $\scriptsize (-3,1)$.
3. Find the length of $\scriptsize \text{DE}$.
4. Find the gradient of $\scriptsize \text{DE}$. Make a conjecture regarding lines $\scriptsize \text{BC}$ and $\scriptsize \text{DE}$.
5. Determine the equation of $\scriptsize \text{BC}$.

The full solutions are at the end of the unit.

# Unit 2: Solutions

### Exercise 2.1

1. Let $\scriptsize \text{A}$ be $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{B}$ be $\scriptsize ({{x}_{2}},{{y}_{2}})$
\scriptsize \begin{align*}{{d}_{{\text{AB}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-6-4)}}^{2}}+{{{(3-(-2))}}^{2}}}}\\&=\sqrt{{{{{(-10)}}^{2}}+{{{(5)}}^{2}}}}\\&=\sqrt{{100+25}}\\&=\sqrt{{125}}\\&=11.2\end{align*}
2. $\scriptsize \text{C}(0,-9)$ be $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{D}(2,3)$ be $\scriptsize ({{x}_{2}},{{y}_{2}})$
\scriptsize \begin{align*}{{d}_{{\text{CD}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(0-2)}}^{2}}+{{{(-9-3)}}^{2}}}}\\&=\sqrt{{{{{(-2)}}^{2}}+{{{(-12)}}^{2}}}}\\&=\sqrt{{4+144}}\\&=\sqrt{{148}}=\sqrt{{4\times 37}}=2\sqrt{{37}}\end{align*}
3. Let $\scriptsize \text{Q}(x,y)$ be $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{T}(x+4,y-7)$ be $\scriptsize ({{x}_{2}},{{y}_{2}})$
\scriptsize \begin{align*}{{d}_{{\text{QT}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(x-(x+4))}}^{2}}+{{{(y-(y-7))}}^{2}}}}\\&=\sqrt{{{{{(4)}}^{2}}+{{{(7)}}^{2}}}}\\&=\sqrt{{16+49}}\\&=\sqrt{{65}}\end{align*}
4. The distance between $\scriptsize \text{S}(0,-3)$ and $\scriptsize \text{F}(8,a)$ is 10 units.
.
Let $\scriptsize \text{S}$ be $\scriptsize ({{x}_{1}},{{y}_{1}})$ and $\scriptsize \text{F}$ be $\scriptsize ({{x}_{2}},{{y}_{2}})$
\scriptsize \begin{align*}{{d}_{{\text{SF}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(0-8)}}^{2}}+{{{(-3-a)}}^{2}}}}\\&=\sqrt{{{{{(-8)}}^{2}}+{{{(-3-a)}}^{2}}}}\\&=\sqrt{{64+{{{(-3-a)}}^{2}}}}\end{align*}
But $\scriptsize {{d}_{{^{{\text{SF}}}}}}=10$
\scriptsize \begin{align*}10 & =\sqrt{{64+{{{(-3-a)}}^{2}}}}\\\therefore {{\left( {10} \right)}^{2}} & =64+{{(-3-a)}^{2}}\quad \text{Expand }{{(-3-a)}^{2}}\\\therefore 100 & =64+9+6a+{{a}^{2}}\quad \text{Solve the quadratic equation}\\\therefore {{a}^{2}}+6a-27 & =0\\\therefore (a+9)(a-3) & =0\\\therefore a=-9\text{ } & \text{or }a=3\end{align*}

Back to Exercise 2.1

### Exercise 2.2

1. .
1. $\scriptsize \text{G}(3,-5)$ and $\scriptsize \text{H}(-2,7)$.
\scriptsize \begin{align*}{{m}_{{\text{GH}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{-5-7}}{{3-(-2)}}\\ & =\displaystyle \frac{{-12}}{5}\end{align*}
2. $\scriptsize \text{G}\left( {\displaystyle \frac{3}{2},0} \right)$ and $\scriptsize \text{H}\left( {3,-\displaystyle \frac{3}{4}} \right)$.
\scriptsize \begin{align*}{{m}_{{\text{GH}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{0-\left( {-\displaystyle \frac{3}{4}} \right)}}{{\displaystyle \frac{3}{2}-3}}\\ & =\displaystyle \frac{{\displaystyle \frac{3}{4}}}{{-\displaystyle \frac{3}{2}}}\\&=\displaystyle \frac{3}{4}\times \left( {-\displaystyle \frac{2}{3}} \right)\\&=-\displaystyle \frac{1}{2}\end{align*}
3. $\scriptsize \text{G}(x-3,y)$ and $\scriptsize \text{H}(x,y-4)$.
\scriptsize \begin{align*}{{m}_{{\text{GH}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{y-(y-4)}}{{(x-3)-x}}\\ & =\displaystyle \frac{{-4}}{{-3}}\\&=\displaystyle \frac{4}{3}\end{align*}
2. $\scriptsize {{m}_{{\text{ST}}}}=\displaystyle \frac{2}{3}$
1. $\scriptsize \text{S}(8,q)$ and $\scriptsize \text{T}(16,2)$.
\scriptsize \begin{align*}{{m}_{{\text{ST}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\\therefore \displaystyle \frac{2}{3} & =\displaystyle \frac{{q-2}}{{8-16}}\\\therefore \displaystyle \frac{2}{3} & =\displaystyle \frac{{q-2}}{{-8}}\\\therefore -16 & =3(q-2)\\\therefore -16 & =3q-6\\\therefore 3q & =-10\\\therefore q & =\displaystyle \frac{{-10}}{3}=-3\displaystyle \frac{1}{3}\end{align*}
2. $\scriptsize \text{S}(3,2q)$ and $\scriptsize \text{T}(9,14)$.
\scriptsize \begin{align*}{{m}_{{\text{ST}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\\therefore \displaystyle \frac{2}{3} & =\displaystyle \frac{{2q-14}}{{3-9}}\\\therefore \displaystyle \frac{2}{3} & =\displaystyle \frac{{2q-14}}{{-6}}\\\therefore -12 & =3(2q-14)\\\therefore -12 & =6q-42\\\therefore 6q & =30\\\therefore q & =\displaystyle \frac{{30}}{6}=5\end{align*}

Back to Exercise 2.2

### Exercise 2.3

1. .
1. $\scriptsize \text{A}(-1,-1)$, $\scriptsize \text{B}(0,-4)$, $\scriptsize \text{C}(3,-4)$, $\scriptsize \text{D}(5,2)$
\scriptsize \begin{align*}{{m}_{{\text{AB}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{-1-(-4)}}{{-1-0}}\\ & =\displaystyle \frac{{-3}}{{-1}}\\ & =3\end{align*}
\scriptsize \begin{align*}{{m}_{{\text{CD}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{-4-2}}{{3-5}}\\ & =\displaystyle \frac{{-6}}{{-2}}\\ & =3\end{align*}
Therefore $\scriptsize \text{AB}$ and $\scriptsize \text{CD}$ are parallel.
2. $\scriptsize \text{A}(-1,3)$, $\scriptsize \text{B}(-2,2)$, $\scriptsize \text{C}(3,-4)$, $\scriptsize \text{D}(5,2)$
\scriptsize \begin{align*}{{m}_{{\text{AB}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{3-2}}{{-1-(-2)}}\\ & =\displaystyle \frac{1}{1}\\ & =1\end{align*}
\scriptsize \begin{align*}{{m}_{{\text{CD}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{-4-2}}{{3-5}}\\ & =\displaystyle \frac{{-6}}{{-2}}\\ & =3\end{align*}
Therefore $\scriptsize \text{AB}$ and $\scriptsize \text{CD}$ are neither parallel nor perpendicular.
2. $\scriptsize \text{K}(-6,2)$, $\scriptsize \text{L}(-3,1)$ and $\scriptsize \text{M}(1,-1)$
\scriptsize \begin{align*}{{m}_{{\text{KL}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{2-1}}{{-6-(-3)}}\\ & =\displaystyle \frac{1}{{-3}}\\ & =-\displaystyle \frac{1}{3}\end{align*}
\scriptsize \begin{align*}{{m}_{{\text{LM}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{1-(-1)}}{{-3-1}}\\ & =\displaystyle \frac{2}{{-4}}\\ & =-\displaystyle \frac{1}{2}\end{align*}
$\scriptsize {{m}_{{\text{KL}}}}\ne {{m}_{{\text{LM}}}}$. Therefore $\scriptsize \text{K}(-6,2)$, $\scriptsize \text{L}(-3,1)$ and $\scriptsize \text{M}(1,-1)$ do not lie on the same straight line.
3. Gradient of the line $\scriptsize 4x-3y=9$:
\scriptsize \begin{align*}4x-3y & =9\\\therefore -3y & =-4x-9\\\therefore y & =\displaystyle \frac{4}{3}x+3\end{align*}
Therefore $\scriptsize {{m}_{{\text{JK}}}}=-\displaystyle \frac{3}{4}$
\scriptsize \begin{align*}{{m}_{{\text{JK}}}} & =-\displaystyle \frac{3}{4}=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\\therefore -\displaystyle \frac{3}{4} & =\displaystyle \frac{{1-(-3)}}{{-3-q}}\\ & =\displaystyle \frac{4}{{-3-q}}\\\therefore -\displaystyle \frac{3}{4} & =\displaystyle \frac{4}{{-3-q}}\\\therefore -3(-3-q) & =16\\\therefore 9+3q & =16\\\therefore 3q & =7\\\therefore q & =\displaystyle \frac{7}{3}\end{align*}

Back to Exercise 2.3

### Exercise 2.4

1. $\scriptsize \text{A}(-4,7)$ and $\scriptsize \text{B}(2,5)$
\scriptsize \begin{align*}\text{M}(x,y) & =\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\\&=\left( {\displaystyle \frac{{-4+2}}{2},\displaystyle \frac{{7+5}}{2}} \right)\\&=\left( {\displaystyle \frac{{-2}}{2},\displaystyle \frac{{12}}{2}} \right)\\&=\left( {-1,6} \right)\end{align*}
2. $\scriptsize \text{G}(x+3,y-2)$ and $\scriptsize \text{H}(x-5,y-4)$
\scriptsize \begin{align*}\text{M}(x,y) & =\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\\&=\left( {\displaystyle \frac{{(x+3)+(x-5)}}{2},\displaystyle \frac{{(y-2)+(y-4)}}{2}} \right)\\&=\left( {\displaystyle \frac{{2x-2}}{2},\displaystyle \frac{{2y-6}}{2}} \right)\\&=\left( {\displaystyle \frac{{2(x-1)}}{2},\displaystyle \frac{{2(y-3)}}{2}} \right)\\&=\left( {x-1,y-3} \right)\end{align*}
3. The midpoint $\scriptsize \text{M}$ of $\scriptsize \text{PQ}$ is $\scriptsize (3,9)$. Find $\scriptsize \text{Q}$ if $\scriptsize \text{P}(1,1)$.
Let $\scriptsize \text{P}(1,1)$ be point $\scriptsize 1$.
x-coordinate:
\scriptsize \begin{align*}3 & =\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2}\\\therefore 3 & =\displaystyle \frac{{1+{{x}_{2}}}}{2}\\\therefore 6 & =1+{{x}_{2}}\\\therefore {{x}_{2}} & =5\end{align*}
.
y-coordinate:
\scriptsize \begin{align*}9 & =\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}\\\therefore 9 & =\displaystyle \frac{{1+{{y}_{2}}}}{2}\\\therefore 18 & =1+{{y}_{2}}\\\therefore {{y}_{2}} & =17\end{align*}
.
$\scriptsize \text{Q}$ is the point $\scriptsize (5,17)$.

Back to Exercise 2.4

### Unit 2: Assessment

1. $\scriptsize \text{A}(-4.5,4)$, $\scriptsize \text{B}(5.5,1.5)$
1. .
\scriptsize \begin{align*}{{d}_{{\text{AB}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-4.5-5.5)}}^{2}}+{{{(4-1.5)}}^{2}}}}\\&=\sqrt{{{{{(-10)}}^{2}}+{{{(2.5)}}^{2}}}}\\&=\sqrt{{100+6.25}}\\&=\sqrt{{106.25}}\\&=10.3\end{align*}
2. .
\scriptsize \begin{align*}{{m}_{{\text{AB}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{1}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{1.5-4}}{{5.5-4.5}}\\ & =\displaystyle \frac{{-2.5}}{1}\\ & =-2\displaystyle \frac{1}{2}=-\displaystyle \frac{5}{2}\end{align*}
3. .
\scriptsize \begin{align*}\text{M}(x,y) & =\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\\&=\left( {\displaystyle \frac{{-4.5+5.5}}{2},\displaystyle \frac{{4+1.5}}{2}} \right)\\&=\left( {\displaystyle \frac{1}{2},\displaystyle \frac{{5.5}}{2}} \right)\\&=\left( {\displaystyle \frac{1}{2},\displaystyle \frac{{\displaystyle \frac{{11}}{2}}}{2}} \right)\\&=\left( {\displaystyle \frac{1}{2},\displaystyle \frac{{11}}{4}} \right)\end{align*}
2. $\scriptsize \text{CD}=5$, $\scriptsize \text{C}(2,4)$, $\scriptsize \text{D}\left( {-1,a} \right)$.
\scriptsize \begin{align*}{{d}_{{\text{CD}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}=5\\\therefore 5 & =\sqrt{{{{{\left( {2-(-1)} \right)}}^{2}}+{{{(4-a)}}^{2}}}}\\ & =\sqrt{{{{3}^{2}}+{{{(4-a)}}^{2}}}}\\ & =\sqrt{{9+{{{(4-a)}}^{2}}}}\\\therefore 25 & =9+{{(4-a)}^{2}}\\\therefore 25 & =9+16-8a+{{a}^{2}}\\\therefore {{a}^{2}}-8a & =0\\\therefore a(a-8) & =0\\\therefore a=0\ & \text{or}\ a=8\end{align*}
We are told that $\scriptsize \text{D}$ has an x-coordinate of $\scriptsize -1$. Therefore $\scriptsize a=0$. $\scriptsize a\ne 8$ in this situation as this would mean that $\scriptsize \text{D}$ has an x-coordinate greater than $\scriptsize 2$.
3. $\scriptsize \text{ABCD}$ is a parellogram where $\scriptsize \text{A}(5,3)$, $\scriptsize \text{B}(2,1)$ and $\scriptsize \text{C}(7,-3)$.
Here is a diagram of the information given.
.

.
We know that the diagonals of a parallelogram bisect each other. Therefore, the midpoint of $\scriptsize \text{AD}$ will be the same point as the midpoint of $\scriptsize \text{BC}$.Midpoint $\scriptsize \text{BC}$
\scriptsize \begin{align*}{{\text{M}}_{{\text{BC}}}} & =\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\\ & =\left( {\displaystyle \frac{{2+7}}{2},\displaystyle \frac{{1-3}}{2}} \right)\\&=\left( {\displaystyle \frac{9}{2},\displaystyle \frac{{-2}}{2}} \right)\\&=\left( {4.5,-1} \right)\end{align*}
$\scriptsize {{\text{M}}_{{\text{AD}}}}=(4.5,-1)$x-coordinate of $\scriptsize \text{D}(x,y)$:
\scriptsize \begin{align*}4.5=\displaystyle \frac{{5+x}}{2}\\\therefore 9=5+x\\\therefore x=4\end{align*}y-coordinate of $\scriptsize \text{D}(x,y)$:
\scriptsize \begin{align*}-1=\displaystyle \frac{{3+y}}{2}\\\therefore -2=3+y\\\therefore y=-5\end{align*}
The coordinates of $\scriptsize \text{D}(x,y)$ are $\scriptsize (4,-5)$.
4. .
1. .
\scriptsize \begin{align*}{{d}_{{\text{EF}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(1-3)}}^{2}}+{{{(5-7)}}^{2}}}}\\&=\sqrt{{{{{(-2)}}^{2}}+{{{(-2)}}^{2}}}}\\&=\sqrt{{4+4}}\\&=\sqrt{8}\\&=2\sqrt{2}\end{align*}
\scriptsize \begin{align*}{{d}_{{\text{FG}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(3-7)}}^{2}}+{{{(7-2)}}^{2}}}}\\&=\sqrt{{{{{(-4)}}^{2}}+{{{(-5)}}^{2}}}}\\&=\sqrt{{16+25}}\\&=\sqrt{{41}}\end{align*}
\scriptsize \begin{align*}{{d}_{{\text{GH}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(7-2)}}^{2}}+{{{(2-0)}}^{2}}}}\\&=\sqrt{{{{{(5)}}^{2}}+{{{(2)}}^{2}}}}\\&=\sqrt{{25+4}}\\&=\sqrt{{29}}\end{align*}
\scriptsize \begin{align*}{{d}_{{\text{HE}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(2-1)}}^{2}}+{{{(0-5)}}^{2}}}}\\&=\sqrt{{{{{(1)}}^{2}}+{{{(-5)}}^{2}}}}\\&=\sqrt{{1+25}}\\&=\sqrt{{26}}\end{align*}
2. .
\scriptsize \begin{align*}{{m}_{{\text{EF}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{1}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{7-5}}{{3-1}}\\ & =\displaystyle \frac{2}{2}\\ & =1\end{align*}
\scriptsize \begin{align*}{{m}_{{\text{FG}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{1}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{2-7}}{{7-3}}\\ & =\displaystyle \frac{{-5}}{4}\end{align*}
\scriptsize \begin{align*}{{m}_{{\text{GH}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{1}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{0-2}}{{2-7}}\\ & =\displaystyle \frac{{-2}}{{-5}}\\&=\displaystyle \frac{2}{5}\end{align*}
\scriptsize \begin{align*}{{m}_{{\text{HE}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{1}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{5-0}}{{1-2}}\\ & =\displaystyle \frac{5}{{-1}}\\&=-5\end{align*}
3. .
\scriptsize \begin{align*}{{\text{M}}_{{\text{EG}}}} & =\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\\ & =\left( {\displaystyle \frac{{1+7}}{2},\displaystyle \frac{{5+2}}{2}} \right)\\&=\left( {\displaystyle \frac{8}{2},\displaystyle \frac{7}{2}} \right)\\&=\left( {4,3.5} \right)\end{align*}
\scriptsize \begin{align*}{{\text{M}}_{{\text{FH}}}} & =\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\\ & =\left( {\displaystyle \frac{{3+2}}{2},\displaystyle \frac{{7+0}}{2}} \right)\\&=\left( {\displaystyle \frac{5}{2},\displaystyle \frac{7}{2}} \right)\\&=\left( {2.5,3.5} \right)\end{align*}
The diagonals do not bisect each other.
4. $\scriptsize \text{EFGH}$ is a quadrilateral. It does not have any equal or parallel sides nor do its diagonals bisect each other.
5. $\scriptsize \text{A}(-2,4)$, $\scriptsize \text{B}(-4,-2)$and $\scriptsize \text{C}(4,0)$ are the vertices of $\scriptsize \Delta \text{ABC}$. $\scriptsize \text{D}$ and $\scriptsize \text{E}(1,2)$ are the midpoints of $\scriptsize \text{AB}$ and $\scriptsize \text{AC}$ respectively.
1. .
\scriptsize \begin{align*}{{m}_{{\text{BC}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{1}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{0-(-2)}}{{4-(-4)}}\\ & =\displaystyle \frac{2}{8}\\&=\displaystyle \frac{1}{4}\end{align*}
2. .
\scriptsize \begin{align*}{{\text{M}}_{{\text{AB}}}} & =\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\\ & =\left( {\displaystyle \frac{{-2-4}}{2},\displaystyle \frac{{4-2}}{2}} \right)\\&=\left( {\displaystyle \frac{{-6}}{2},\displaystyle \frac{2}{2}} \right)\\&=\left( {-3,1} \right)\end{align*}
$\scriptsize \text{D}$ is the midpoint of $\scriptsize \text{AB}$. Therefore $\scriptsize \text{D}(-3,1)$.
3. .
\scriptsize \begin{align*}{{d}_{{\text{DE}}}} & =\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-3-1)}}^{2}}+{{{(1-2)}}^{2}}}}\\&=\sqrt{{{{{(-4)}}^{2}}+{{{(-1)}}^{2}}}}\\&=\sqrt{{16+1}}\\&=\sqrt{{17}}\end{align*}
4. .
\scriptsize \begin{align*}{{m}_{{\text{DE}}}} & =\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{1}}-{{x}_{1}}}}\\ & =\displaystyle \frac{{2-1}}{{1-(-3)}}\\ & =\displaystyle \frac{1}{4}\end{align*}
$\scriptsize {{m}_{{\text{BC}}}}={{m}_{{\text{DE}}}}$ and $\scriptsize \text{D}$ and $\scriptsize \text{E}$ are the midpoints of $\scriptsize \text{AB}$ and $\scriptsize \text{AC}$ respectively. Therefore, the length of $\scriptsize \text{DE}$ is half the length of $\scriptsize \text{AB}$.
5. .
$\scriptsize {{m}_{{\text{BC}}}}=\displaystyle \frac{1}{4}$. Therefore, the equation of $\scriptsize \text{BC}$ is of the form $\scriptsize y=\displaystyle \frac{1}{4}x+c$. But $\scriptsize \text{C}(4,0)$ is a point on $\scriptsize \text{BC}$. Substitute these coordinates into the equation to solve for $\scriptsize c$. \scriptsize \begin{align*}0 & =\displaystyle \frac{1}{4}(4)+c\\\therefore 0 & =1+c\\\therefore c & =-1\end{align*}
Therefore, the equation of $\scriptsize \text{BC}$ is $\scriptsize y=\displaystyle \frac{1}{4}x-1$.

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