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You are here: Home Grade 11 Mathematics Analytical geometry Perpendicular lines

Perpendicular lines

Activity 1: Investigation: Perpendicular lines

  1. 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 \(D(4;-2)\).
  2. Label and measure \(\alpha\) and \(\beta\), the angles of inclination of straight lines \(AB\) and \(CD\) respectively.
  3. Label and measure \(\theta\), the angle between the lines \(AB\) and \(CD\).
  4. Describe the relationship between the lines \(AB\) and \(CD\).
  5. “\(\theta\) is a reflex angle, therefore \(AB \perp CD\).” Is this a true statement? If not, provide a correct statement.
  6. Determine the equation of the straight line \(AB\) and the line \(CD\).
  7. Use your calculator to determine \(\tan \alpha \times \tan \beta\).
  8. Determine \(m_{AB} \times m_{CD}\).
  9. What do you notice about these products?
  10. Complete the sentence: if two lines are \(\ldots \ldots\) to each other, then the product of their \(\ldots \ldots\) is equal \(\ldots \ldots\)
  11. 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: \(m_1 \times m_2 = -1\)

Image

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 \(\textrm{90}\text{°}\) and place point \(B\) at \((-3;4)\) so that we have the angle of inclination \(B\hat{O}X = \textrm{90}\text{°}+ \theta\).

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 \(\textrm{90}\text{°}\) 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

\[m_{OA} = -\frac{1}{m_{OB}}\]
Image

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 \(B\hat{O}X = \textrm{90}\text{°}+ \theta\), 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 \(y = m_2x + c_2\).

Image

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.

Example 1: Perpendicular lines

Question

Determine the equation of the straight line passing through the point \(T(2;2)\) and perpendicular to the line \(3y + 2x - 6 = 0\).

Answer

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 \(T(2;2)\).

\[\begin{align} y - 2 & = \frac{3}{2}(x - 2) \\ y - 2& = \frac{3}{2}x - 3\\ y & = \frac{3}{2}x - 1 \end{align}\]
Image

A sketch was not required, but it is useful for checking the answer.

Write the final answer

The equation of the straight line is \(y = \frac{3}{2}x - 1\).

Example 2: Perpendicular lines

Question

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 71,57°.

Answer

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 \(m_2\).

\[\begin{align*} m_{2} & = \tan \theta \\ & = \tan \textrm{71,57}\text{°} \\ & = \textrm{3,0} \end{align*}\]

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 \(y = -\frac{1}{3}x + 1\).

Exercise 1: Perpendicular lines

Calculate whether or not the following two lines are perpendicular:

  1. \(y - 1 = 4x\) and \(4y + x + 2 = 0\)

  2. \(10x = 5y - 1\) and \(5y - x - 10 = 0\)

  3. \(x = y - 5\) and the line passing through \((-1;\frac{5}{4})\) and \((3;-\frac{11}{4})\)

  4. \(y = 2\) and \(x = 1\)

  5. \(\frac{y}{3} = x\) and \(3y + x = 9\)

  6. \(1 - 2x = y\) and the line passing through \((2;-1)\) and \((-1;5)\)

  7. \(y = x + 2\) and \(2y + 1 = 2x\)

  1. \[\begin{aligned} y - 1 &= 4x \\ y &= 4x + 1 \\ \therefore m_1 &= 4\\ 4y + x + 2 &= 0 \\ 4y &= - x - 2 \\ y &= - \frac{1}{4}x - \frac{1}{2} \\ \therefore m_2 &= -\frac{1}{4} \\ \therefore m_1 \times m_2&= -\frac{1}{4} \times 4 \\ &= -1 \end{aligned}\]

    \(\therefore\)Perpendicular lines.

  2. \[\begin{aligned} 10x &= 5y - 1 \\ 10x + 1 &= 5y \\ 2x + \frac{1}{5} &= y \\ \therefore m_1 &= 2 \\ 5y - x - 10 &= 0 \\ 5y &= - x + 10 \\ y &= - \frac{1}{5}x + 2 \\ \therefore m_2 &= -\frac{1}{5} \\ \therefore m_1 \times m_2&= -\frac{1}{5} \times 2 \\ &\ne -1 \end{aligned}\]

    \(\therefore\)Not perpendicular lines.

  3. \[\begin{aligned} x &= y - 5 \\ x + 5 &= y \\ \therefore m_1 &= 1 \\ m_2 &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{-\frac{11}{4} -\frac{5}{4}}{3 + 1} \\ &= \frac{-4}{4} \\ \therefore m_2 &= -1 \\ \therefore m_1 \times m_2&= -1 \times 1 \\ &= -1 \end{aligned}\]

    \(\therefore\)Perpendicular lines.

  4. Perpendicular: horizontal and vertical lines.
  5. \[\begin{aligned} \frac{y}{3} &= x \\ y &= 3x \\ \therefore m_1 &= 3 \\ 3y + x &= 9 \\ 3y &= -x + 9 \\ y &= -\frac{1}{3}x + 3 \\ \therefore m_2 &= -\frac{1}{3} \\ \therefore m_1 \times m_2&= -\frac{1}{3} \times 3 \\ &= -1 \end{aligned}\]

    \(\therefore\)Perpendicular lines.

  6. \[\begin{aligned} 1 -2x &= y \\ \therefore m_1 &= -2 \\ m_2 &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{5 + 1}{-1-2} \\ &= \frac{6}{-3} \\ \therefore m_2 &= -2 \\ \therefore m_1 \times m_2&= -2 \times -2 \\ &\ne -1 \end{aligned}\]

    \(\therefore\)Not perpendicular lines.

  7. \[\begin{aligned} y &= x + 2 \\ \therefore m_1 &= 1 \\ 2y + 1 &= 2x \\ 2y &= 2x - 1 \\ y &= x - \frac{1}{2} \\ \therefore m_2 &= 1 \\ \therefore m_1 \times m_2&= 1 \times 1 \\ &\ne -1 \end{aligned}\]

    \(\therefore\)Not perpendicular lines.

Determine the equation of the straight line that passes through the point \((-2;-4)\) and is perpendicular to the line \(y + 2x = 1\).

\[\begin{aligned} y &= -2x + 1 \\ \therefore m_1 &= -2 \\ \text{For } \perp: m_1 \times m_2 &= -1 \\ -2 \times m_2 &= -1 \\ \therefore m_2 &= \frac{1}{2} \\ y &= mx + c \\ y &= \frac{1}{2}x + c \\ \text{Subst. } (-2;-4): \quad - 4 &= \frac{1}{2}(-2) + c \\ -4 &= -1 + c \\ \therefore c &= -3 \\ \therefore y &= \frac{1}{2}x - 3 \end{aligned}\]

Determine the equation of the straight line that passes through the point \((2;-7)\) and is perpendicular to the line \(5y - x = 0\).

\[\begin{aligned} 5y - x &= 0 \\ 5y &= x \\ y &= \frac{1}{5}x \\ \therefore m_1 &= \frac{1}{5} \\ \text{For } \perp: m_1 \times m_2 &= -1 \\ \frac{1}{5} \times m_2 &= -1 \\ \therefore m_2 &= -5 \\ y &= mx + c \\ y &= -5x + c \\ \text{Subst. } (2;7): \quad -7 &= -5(2) + c \\ -7 &= -10 + c \\ \therefore c &= 3 \\ \therefore y &= -5x + 3 \end{aligned}\]

Determine the equation of the straight line that passes through the point \((3;-1)\) and is perpendicular to the line with angle of inclination \(\theta = \textrm{135}\text{°}\).

\[\begin{align*} \theta &= \textrm{135}\text{°} \\ \therefore m &= \tan \theta \\ &= \tan \textrm{135}\text{°} \\ \therefore m &= -1 \\ y &= mx + c \\ y &= -x + c \\ \text{Subst. } (3;-1): \quad -1 &= -(3) + c \\ \therefore c &= 2 \\ \therefore y &= -x + 2 \end{align*}\]

Determine the equation of the straight line that passes through the point \((-2;\frac{2}{5})\) and is perpendicular to the line \(y = \frac{4}{3}\).