# What is the equation of the line that is perpendicular to the line passing through (3,18) and (6,14) at midpoint of the two points?

$6 x - 8 y + 121 = 0$

#### Explanation:

The slope $m$ of line perpendicular to the line joining the points $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(3 , 18\right)$ & $\left({x}_{2} , {y}_{2}\right) \setminus \equiv \left(6 , 14\right)$ is given as

$m = - \frac{1}{\setminus \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}}$

$= - \frac{1}{\setminus \frac{14 - 18}{6 - 3}}$

$= \frac{3}{4}$

Now, the mid-point of line joining the points $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(3 , 18\right)$ & $\left({x}_{2} , {y}_{2}\right) \setminus \equiv \left(6 , 14\right)$ are given as

$\left(\setminus \frac{{x}_{1} + {x}_{2}}{2} , \setminus \frac{{y}_{1} + {y}_{2}}{2}\right)$

$\setminus \equiv \left(\setminus \frac{3 + 6}{2} , \setminus \frac{18 + 14}{2}\right)$

$\setminus \equiv \left(\setminus \frac{9}{2} , 16\right)$

Now, the equation of line passing through the point $\left({x}_{0} , {y}_{0}\right) \setminus \equiv \left(\frac{9}{2} , 16\right)$ & having slope $m = \frac{3}{4}$ is given as

$y - {y}_{0} = m \left(x - {x}_{0}\right)$

$y - 16 = \frac{3}{4} \left(x - \frac{9}{2}\right)$

$6 x - 8 y + 121 = 0$