### 1 Perpendicular lines

One form for the equation of a straight line is

$\phantom{\rule{2em}{0ex}}y=mx+c$

where $m$ and $c$ are constants. We remember that $m$ is the gradient of the line and its value is the tangent of the angle $\theta $ that the line makes with the positive $x$ -axis. The constant $c$ is the value obtained where the line intersects the $y$ -axis. See Figure 1:

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Figure 1
**

If we have a second line, with equation

$\phantom{\rule{2em}{0ex}}y=nx+d$

then, unless $m=n$ , the two lines will intersect at one point. These are drawn together in Figure 2. The second line makes an angle $\psi $ with the positive $x$ -axis.

**
Figure 2
**

A simple question to ask is “what is the relation between $m$ and $n$ if the lines are perpendicular?” If the lines are perpendicular, as shown in Figure 3, the angles $\theta $ and $\psi $ must satisfy the relation:

$\phantom{\rule{2em}{0ex}}\psi -\theta =9{0}^{\circ}$

**
Figure 3
**

This is true since the angles in a triangle add up to $18{0}^{\circ}$ . According to the figure the three angles are $9{0}^{\circ}$ , $\theta $ and $18{0}^{\circ}-\psi $ . Therefore

$\phantom{\rule{2em}{0ex}}18{0}^{\circ}=9{0}^{\circ}+\theta +\left(18{0}^{\circ}-\psi \right)\phantom{\rule{2em}{0ex}}\text{implying}\phantom{\rule{2em}{0ex}}\psi -\theta =9{0}^{\circ}$

In this special case that the lines are perpendicular or
**
normal
**
to each other the relation between the gradients
$m$
and
$n$
is easily obtained. In this deduction we use the following basic trigonometric relations and identities:

$\phantom{\rule{2em}{0ex}}sin\left(A-B\right)\equiv sinAcosB-cosAsinB\phantom{\rule{2em}{0ex}}cos\left(A-B\right)\equiv cosAcosB+sinAsinB$

$\phantom{\rule{2em}{0ex}}tanA\equiv \frac{sinA}{cosA}\phantom{\rule{2em}{0ex}}sin9{0}^{\circ}=1\phantom{\rule{2em}{0ex}}cos9{0}^{\circ}=0$

Now

$$\begin{array}{rcll}m& =& tan\theta & \text{}\\ & =& tan\left(\psi -9{0}^{o}\right)\phantom{\rule{2em}{0ex}}\text{(seeFigure3)}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& \frac{sin\left(\psi -9{0}^{o}\right)}{cos\left(\psi -9{0}^{o}\right)}& \text{}\\ & =& \frac{-cos\psi}{sin\psi}=-\frac{1}{tan\psi}=-\frac{1}{n}& \text{}\\ \text{So}\phantom{\rule{2em}{0ex}}mn& =& -1& \text{}\end{array}$$##### Key Point 1

Two straight lines $y=mx+c$ , $y=nx+d$ are perpendicular if

##### Exercise

Which of the following pairs of lines are perpendicular?

- $y=-x+1,\phantom{\rule{1em}{0ex}}y=x+1$
- $y+x-1=0,\phantom{\rule{1em}{0ex}}y+x-2=0$
- $2y=8x+3,\phantom{\rule{1em}{0ex}}y=-0.25x-1$

- perpendicular
- not perpendicular
- perpendicular