4.5 Perpendicular lines (EMBGG)
Perpendicular lines
- Draw a sketch of the line passing through the points \(A(-2;-3)\) and \(B(2;5)\) and the line passing
through the points \(C(-1;\frac{1}{2})\) and \(D(4;-2)\).
- Label and measure \(\alpha\) and \(\beta\), the angles of inclination of straight lines \(AB\) and \(CD\)
respectively.
- Label and measure \(\theta\), the angle between the lines \(AB\) and \(CD\).
- Describe the relationship between the lines \(AB\) and \(CD\).
- “\(\theta\) is a reflex angle, therefore \(AB \perp CD\).” Is this a true statement? If not,
provide a correct statement.
- Determine the equation of the straight line \(AB\) and the line \(CD\).
- Use your calculator to determine \(\tan \alpha \times \tan \beta\).
- Determine \(m_{AB} \times m_{CD}\).
- What do you notice about these products?
- Complete the sentence: if two lines are \(\ldots \ldots\) to each other, then the product of their \(\ldots
\ldots\) is equal \(\ldots \ldots\)
- Complete the sentence: if the gradient of a straight line is equal to the negative \(\ldots \ldots\) of the
gradient of another straight line, then the two lines are \(\ldots \ldots\)
Deriving the formula: \(m_1 \times m_2 = -1\)
Consider the point \(A(4;3)\) on the Cartesian plane with an angle of inclination \(A\hat{O}X = \theta\). Rotate
through an angle of \(\text{90}\text{°}\) and place point \(B\) at \((-3;4)\) so that we have the angle of
inclination \(B\hat{O}X = \text{90}\text{°}+ \theta\).
We determine the gradient of \(OA\):
\begin{align*}
m_{OA} &= \frac{y_2 - y_1}{x_2 - x_1} \\
&= \frac{3 - 0}{4 - 0} \\
&= \frac{3}{4}
\end{align*}
And determine the gradient of \(OB\):
\begin{align*}
m_{OB} &= \frac{y_2 - y_1}{x_2 - x_1} \\
&= \frac{4 - 0}{-3 - 0} \\
&= \frac{4}{-3}
\end{align*}
By rotating through an angle of \(\text{90}\text{°}\) we know that \(OB \perp OA\):
\begin{align*}
m_{OA} \times m_{OB} &= \frac{3}{4} \times \frac{4}{-3} \\
&= -1
\end{align*}
We can also write that
\[m_{OA} = -\frac{1}{m_{OB}}\]
If we have the general point \(A(x;y)\) with an angle of inclination \(A\hat{O}X = \theta\) and point \(B(-y;x)\)
such that \(B\hat{O}X = \text{90}\text{°}+ \theta\), then we know that
\begin{align*}
m_{OA} &= \frac{y}{x} \\
m_{OB} &= -\frac{x}{y} \\
\therefore m_{OA} \times m_{OB}&= \frac{y}{x} \times -\frac{x}{y} \\
&= -1
\end{align*}
Another method of determining the equation of a straight line is to be given a point on the line,
\(\left({x}_{1};{y}_{1}\right)\), and the equation of a line which is perpendicular to the unknown line. Let the
equation of the unknown line be \(y = m_1x + c_1\) and the equation of the given line be \(y = m_2x + c_2\).
If the two lines are perpendicular then
\[m_1 \times m_2 = -1\]
Note: this rule does not apply to vertical or horizontal lines.
When determining the gradient of a line using the coefficient of \(x\), make sure the given equation is written
in the gradient–intercept (standard) form \(y = mx + c\). Then we know that \[m_1 = -\frac{1}{m_2}\]
Substitute the value of \(m_1\) and the given point \(\left({x}_{1};{y}_{1}\right)\), into the
gradient–intercept form of the straight line equation \(y-y_1 = m(x-x_1)\) and determine the equation of the
unknown line.
Worked example 13: Perpendicular lines
Determine the equation of the straight line passing through the point \(T(2;2)\) and perpendicular to the
line \(3y + 2x - 6 = 0\).
Write the equation in standard form
Let the gradient of the unknown line be \(m_1\) and the given gradient be \(m_2\). We write the given
equation in gradient–intercept form and determine the value of \(m_2\).
\begin{align*}
3y + 2x - 6 &= 0 \\
3y & = -2x + 6 \\
y & = -\frac{2}{3}x + 2\\
\therefore m_2 & = -\frac{2}{3}
\end{align*}
We know that the two lines are perpendicular, therefore \(m_1 \times m_2 = -1\). Therefore \(m_1 =
\frac{3}{2}\).
Write down the gradient–point form of the straight line equation
\[y - y_1 = m(x - x_1)\]
Substitute \(m_1 = \frac{3}{2}\).
\[y - y_1 = \frac{3}{2}(x - x_1)\]
Substitute the given point \(T(2;2)\).
\begin{align*}
y - 2 & = \frac{3}{2}(x - 2) \\
y - 2& = \frac{3}{2}x - 3\\
y & = \frac{3}{2}x - 1
\end{align*}
A sketch was not required, but it is useful for checking the answer.
Write the final answer
The equation of the straight line is \(y = \frac{3}{2}x - 1\).
Worked example 14: Perpendicular lines
Determine the equation of the straight line passing through the point \((2;\frac{1}{3})\) and perpendicular
to the line with an angle of inclination of \(\text{71,57}\)\(\text{°}\).
Use the given angle of inclination to determine gradient
Let the gradient of the unknown line be \(m_1\) and let the given gradient be \(m_2\).
\begin{align*}
m_{2} & = \tan \theta \\
& = \tan \text{71,57}\text{°} \\
& = \text{3,0}
\end{align*}
Determine the unknown gradient
Since we are given that the two lines are perpendicular,
\begin{align*}
m_{1} \times m_{2} &= -1 \\
\therefore m_{1} &= - \frac{1}{3}
\end{align*}
Write down the gradient–point form of the straight line equation
\[y - y_1 = m(x - x_1)\]
Substitute the gradient \(m_{1} = -\frac{1}{3}\).
\[y - y_1 = -\frac{1}{3}(x - x_1)\]
Substitute the given point \((2;\frac{1}{3})\).
\begin{align*}
y - \left(\frac{1}{3}\right) & = -\frac{1}{3}(x - 2) \\
y -\frac{1}{3} & = -\frac{1}{3}x + \frac{2}{3}\\
y & = -\frac{1}{3}x + 1
\end{align*}
Write the final answer
The equation of the straight line is \(y = -\frac{1}{3}x + 1\).
Perpendicular lines
Textbook Exercise 4.8
\(y - 1 = 4x\) and \(4y + x + 2 = 0\)
\begin{align*}
y - 1 &= 4x \\
y &= 4x + 1 \\
\therefore m_1 &= 4\\
4y + x + 2 &= 0 \\
4y &= - x - 2 \\
y &= - \frac{1}{4}x - \frac{1}{2} \\
\therefore m_2 &= -\frac{1}{4} \\
\therefore m_1 \times m_2&= -\frac{1}{4} \times 4 \\
&= -1
\end{align*}
\(\therefore\)Perpendicular lines.
\(10x = 5y - 1\) and \(5y - x - 10 = 0\)
\begin{align*}
10x &= 5y - 1 \\
10x + 1 &= 5y \\
2x + \frac{1}{5} &= y \\
\therefore m_1 &= 2 \\
5y - x - 10 &= 0 \\
5y &= - x + 10 \\
y &= - \frac{1}{5}x + 2 \\
\therefore m_2 &= -\frac{1}{5} \\
\therefore m_1 \times m_2&= -\frac{1}{5} \times 2 \\
&\ne -1
\end{align*}
\(\therefore\)Not perpendicular lines.
\(x = y - 5\) and the line passing through \((-1;\frac{5}{4})\) and \((3;-\frac{11}{4})\)
\begin{align*}
x &= y - 5 \\
x + 5 &= y \\
\therefore m_1 &= 1 \\
m_2 &= \frac{y_2 - y_1}{x_2 - x_1} \\
&= \frac{-\frac{11}{4} -\frac{5}{4}}{3 + 1} \\
&= \frac{-4}{4} \\
\therefore m_2 &= -1 \\
\therefore m_1 \times m_2&= -1 \times 1 \\
&= -1
\end{align*}
\(\therefore\)Perpendicular lines.
Perpendicular: horizontal and vertical lines.
\(\frac{y}{3} = x\) and \(3y + x = 9\)
\begin{align*}
\frac{y}{3} &= x \\
y &= 3x \\
\therefore m_1 &= 3 \\
3y + x &= 9 \\
3y &= -x + 9 \\
y &= -\frac{1}{3}x + 3 \\
\therefore m_2 &= -\frac{1}{3} \\
\therefore m_1 \times m_2&= -\frac{1}{3} \times 3 \\
&= -1
\end{align*}
\(\therefore\)Perpendicular lines.
\(1 - 2x = y\) and the line passing through \((2;-1)\) and \((-1;5)\)
\begin{align*}
1 -2x &= y \\
\therefore m_1 &= -2 \\
m_2 &= \frac{y_2 - y_1}{x_2 - x_1} \\
&= \frac{5 + 1}{-1-2} \\
&= \frac{6}{-3} \\
\therefore m_2 &= -2 \\
\therefore m_1 \times m_2&= -2 \times -2 \\
&\ne -1
\end{align*}
\(\therefore\)Not perpendicular lines.
\(y = x + 2\) and \(2y + 1 = 2x\)
\begin{align*}
y &= x + 2 \\
\therefore m_1 &= 1 \\
2y + 1 &= 2x \\
2y &= 2x - 1 \\
y &= x - \frac{1}{2} \\
\therefore m_2 &= 1 \\
\therefore m_1 \times m_2&= 1 \times 1 \\
&\ne -1
\end{align*}
\(\therefore\)Not perpendicular lines.
Determine the equation of the straight line that passes through the point \((-2;-4)\) and is
perpendicular to the line \(y + 2x = 1\).
\begin{align*}
y &= -2x + 1 \\
\therefore m_1 &= -2 \\
\text{For } \perp: m_1 \times m_2 &= -1 \\
-2 \times m_2 &= -1 \\
\therefore m_2 &= \frac{1}{2} \\
y &= mx + c \\
y &= \frac{1}{2}x + c \\
\text{Subst. } (-2;-4): \quad - 4 &= \frac{1}{2}(-2) + c \\
-4 &= -1 + c \\
\therefore c &= -3 \\
\therefore y &= \frac{1}{2}x - 3
\end{align*}
Determine the equation of the straight line that passes through the point \((2;-7)\) and is perpendicular
to the line \(5y - x = 0\).
\begin{align*}
5y - x &= 0 \\
5y &= x \\
y &= \frac{1}{5}x \\
\therefore m_1 &= \frac{1}{5} \\
\text{For } \perp: m_1 \times m_2 &= -1 \\
\frac{1}{5} \times m_2 &= -1 \\
\therefore m_2 &= -5 \\
y &= mx + c \\
y &= -5x + c \\
\text{Subst. } (2;7): \quad -7 &= -5(2) + c \\
-7 &= -10 + c \\
\therefore c &= 3 \\
\therefore y &= -5x + 3
\end{align*}
Determine the equation of the straight line that passes through the point \((3;-1)\) and is perpendicular
to the line with angle of inclination \(\theta = \text{135}\text{°}\).
\begin{align*}
\theta &= \text{135}\text{°} \\
\therefore m &= \tan \theta \\
&= \tan \text{135}\text{°} \\
\therefore m_{1} &= -1 \\
\text{For } \perp: m_{1} \times m_{2} &= -1 \\
-1 \times m_{2} &= -1 \\
\therefore m_{2} &= 1 \\
y &= mx + c \\
y &= x + c \\
\text{Subst. } (3;-1): \quad -1 &= (3) + c \\
\therefore c &= -4 \\
\therefore y &= x - 4
\end{align*}
Determine the equation of the straight line that passes through the point \((-2;\frac{2}{5})\) and is
perpendicular to the line \(y = \frac{4}{3}\).
\(x = -2\)
