# What is the equation of the line that is perpendicular to the line passing through (3,18) and (-5,12) at midpoint of the two points?

Sep 5, 2016

$4 x + 3 y - 41 = 0$

#### Explanation:

There could be two ways.

One - The midpoint of $\left(3 , 18\right)$ and $\left(- 5 , 12\right)$ is $\left(\frac{3 - 5}{2} , \frac{18 + 12}{2}\right)$ or $\left(- 1 , 15\right)$.

The slope of line joining $\left(3 , 18\right)$ and $\left(- 5 , 12\right)$ is $\frac{12 - 18}{- 5 - 3} = - \frac{6}{-} 8 = \frac{3}{4}$

Hence, slope of line perpendicular to it will be $- \frac{1}{\frac{3}{4}} = - \frac{4}{3}$ and equation of line passing through $\left(- 1 , 15\right)$ and having a slope of $- \frac{4}{3}$ is

$\left(y - 15\right) = - \frac{4}{3} \left(x - \left(- 1\right)\right)$ or

$3 y - 45 = - 4 x - 4$ or

$4 x + 3 y - 41 = 0$

Two - A line which is perpendicular to line joining $\left(3 , 18\right)$ and $\left(- 5 , 12\right)$ and passes through their midpoint is locus of a point which is equidistant from these two points. Hence, equation is

${\left(x - 3\right)}^{2} + {\left(y - 18\right)}^{2} = {\left(x + 5\right)}^{2} + {\left(y - 12\right)}^{2}$ or

${x}^{2} - 6 x + 9 + {y}^{2} - 36 y + 324 = {x}^{2} + 10 x + 25 + {y}^{2} - 24 y + 144$ or

$- 6 x - 10 x - 36 y + 24 y + 333 - 169 = 0$ or

$- 16 x - 12 y + 164 = 0$ and dividing by $- 4$, we get

$4 x + 3 y - 41 = 0$

Sep 5, 2016

$4 x + 3 y = 41$.

#### Explanation:

The Mid-point M of the segment joining $A \left(3 , 18\right) \mathmr{and} B \left(- 5 , 12\right)$ is

$M \left(\frac{- 5 + 3}{2} , \frac{12 + 18}{2}\right) = M \left(- 1 , 15\right)$

Slope of Line $A B$ is $\frac{18 - 12}{3 - \left(- 5\right)} = \frac{6}{8} = \frac{3}{4}$

Therefore, the slope of the line $\bot \text{ to line } A B = - \frac{4}{3}$

Thus, the reqd. line has slope$= - \frac{4}{3} \text{, and, it passes thro. pt. } M$.

Using, the Slope-Point Form , the reqd. line is :

# y-15=-4/3(x+1), i.e., 3y-45+4x+4=0, or,

$4 x + 3 y = 41$.