**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 $2y – 3x$ = $1$