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, $2(y - 1)$ = $3(x - 1)$
$2y – 2$ = $3x – 3$
$2y$ = $3x + 1$
$2y – 3x$ = $1$
Thus, the equation of the perpendicular bisector is $2y – 3x$ = $1$