Perpendicular Equations

Two straight lines are said to be perpendicular to each other if they intersect at right angle. The angle formed at the point of intersection of the two perpendicular lines is $90$ degrees. When a vertical line cuts a horizontal line then these two lines are said to be perpendicular to each other. The relation between the slopes of two perpendicular lines is that the product of the slopes of the two perpendicular lines is always $-1$. If slope or gradient of one of the lines is known to us then we can solve for the gradient of its perpendicular line by taking negative of the reciprocal of the slope given.

Equation of a perpendicular line can be derived only if the equation of the other perpendicular line or the slope of the line to which it is perpendicular to is given. When asked to find out the equation of a perpendicular line we first arrange the equation of the line to which it is perpendicular to in the slope intercept form $y$ = $mx + c$, where $m$ is the slope. We then find out the slope of the perpendicular line by taking reciprocal of that slope and negative of that value. That is slope of the perpendicular line would be $\frac{-1}{m}$. Having known coordinates of any point lying on the perpendicular line, we plugin the values of those coordinates and slope of the perpendicular line in the slope intercept form and solve for the $y$ intercept $c$. At the end we plugin the value of the slope $\frac{-1}{m}$ and $y$ intercept ‘$c$’ in the slope intercept form $y$ = $mx + c$ and form the equation of the perpendicular line.

The relation between slopes of two perpendicular lines is as follows:

$m_1 \times m_2$ = $-1$

Having known slope of any line say m_1, slope of the perpendicular line can be found out by $m_2$ = $\frac{-1}{m_1}$. We need to know the coordinates of any point lying on the perpendicular line so that we can plugin those values of $x$ and $y$ in the slope intercept form $y$ = $mx + b$. We also plugin the value of the gradient ‘$m$’. This is done to find out the $y$ intercept ‘$b$’. On finding the value of both slope ‘$m$’ and $y$ intercept ‘$b$’ we substitute those values in the slope intercept form of linear equation to form the equation of the perpendicular line.

Example 1:

Find the slope of a line which is perpendicular to the line $2x + 5y$ = $8$

Solution:

To find out the slope of the perpendicular line, we first find out the slope of the given line. We arrange the given equation in slope intercept form $y$ = $mx + b$

$2x + 5y$ = $8$

Separating out $y$ from the given equation step wise.

Step 1: Subtract $2x$ from both sides, $5y$ = $8 – 2x$

Step 2: Divide both sides by $5,\ y$ = $\frac{-2x}{5}$ + $\frac{8}{5}$

Step 3: Identify the value of the slope. Slope $(m1_)$ = $\frac{-2}{5}$

Next we find out the slope of the perpendicular line by using the relation between the slope of two perpendicular lines $m_1 \times m_2$ = $-1$

$\frac{-2}{5}$ $\times m_2$ = $-1$

$m_2$ = $\frac{-1}{(\frac{-2}{5})}$

$m_2$ = $\frac{5}{2}$

Thus the slope of the perpendicular line is $\frac{5}{2}$

Example 2:

Two lines $l_1$ and $l_2$ are perpendicular to each other. Slope of line $l_1$ = $\frac{2}{3}$ and slope of line $l_2$ = $\frac{8}{x}$ - $6$. Find out the value of $x$

Solution:

Using the relation between the slope of two perpendicular lines $m_1 \times m_2$ = $-1$ we are going to find out the value of $x$. Substitute the values of the slope of two lines $11$ and $12$ in the relation,

$m_1 \times m_2$ = $-1$

$\frac{2}{3}$ $\times$ $\frac{8}{x - 6}$ = $-1$

$\frac{16}{3}$ $(x - 6)$ = $-1$

By cross multiplication,

$16$ = $-3(x - 6)$

$16$ = $-3x + 18$

Subtracting $18$ from both sides,

$16 – 18$ = $-3x$

$-2$ = $-3x$

$x$ = $\frac{2}{3}$

Thus, the value of $x$ we get is $\frac{2}{3}$

Example 3:

Find the equation of the line which is perpendicular to the line $y$ = $3x + 2$ and passes through the point $(7, 2)$

Solution:

The slope of the given line is $3$. Therefore, the slope of the line perpendicular to the given line is the negative reciprocal to it. That is
$m_2$ = $\frac{-1}{m_1}$, $m_2$ = $\frac{-2}{m_1}$ = $\frac{-1}{3}$. We are also given the coordinates of a point which lies on the perpendicular line. We plugin the values of the coordinates and the slope in the slope intercept form $y$ = $mx + b$ to solve the $y$ intercept ‘$b$’. So,

$y$ = $mx + b$

$2$ = $-\frac{1}{3 \times 7}$ + $b$

$2$ = $\frac{-7}{3}$ + $b$

$b$ = $2$ + $\frac{7}{3}$

$b$ = $\frac{6}{3}$ + $\frac{7}{3}$

$b$ = $\frac{13}{3}$

On finding out the value of both gradient ‘$m$’ and $y$ intercept ‘$b$’ we now substitute these two values in the slope intercept form

$y$ = $mx + b$

$y$ = $(\frac{-1}{3})$ $\times x$ + $\frac{13}{3}$

$y$ = $\frac{-x}{3}$ + $\frac{13}{3}$

$y$ = $\frac{(-x + 13)}{3}$

$3y$ = $-x + 13$

$3y + x$ = $13$

Thus, the equation of the perpendicular line is $3y + x$ = $13$

Example 4:

Find the equation of the perpendicular bisector of the line segment whose endpoints are $(-2, 3)$ and $(4, -1)$

Solution:

Perpendicular bisector always passes through the midpoint of the line segment. Given the coordinates of the endpoint we find out the coordinates of the midpoint using midpoint formula $x$ = $\frac{(x_1 + x_2)}{2}$ and $y$ = $\frac{(y_1 + y_2)}{2}$

$x$ = $\frac{(-2 + 4)}{2}$ = $1$

$y$ = $\frac{(3 – 1)}{2}$ = $1$

We also find out the slope of the line segment using the slope formula $m$ = $\frac{(y_2 – y_1)}{(x_2 – x_1)}$

Given, $x_1$ = $-2,\ y_1$ = $3,\ x_2$ = $4,\ y_2$ = $-1$

$m$ = $\frac{(-1 -3)}{(4 + 2)}$ = $\frac{-2}{3}$

Slope of the perpendicular line is the negative reciprocal. So gradient $(m_2)$ = $-(\frac{1}{(\frac{-2}{3})})$ = $\frac{3}{2}$

Using point slope form $(y – y_1)$ = $m\ (x – x_1)$

We substitute the values of $x_1$ = $1,\ y_1$ = $1,\ m$ = $\frac{3}{2}$,

$(y - 1)$ = $\frac{3}{2(x – 1)}$

On cross multiplication, $2(y - 1)$ = $3(x - 1)$

$2y – 2$ = $3x – 3$

$2y$ = $3x + 1$