# What is the slope of any line perpendicular to the line passing through (3,-2) and (12,19)?

Feb 10, 2016

Slope of any line perpendicular to the line passing through (3,−2) and $\left(12 , 19\right)$ is $- \frac{3}{7}$

#### Explanation:

If the two points are $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$, the slope of the line joining them is defined as

$\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ or $\frac{{y}_{1} - {y}_{2}}{{x}_{1} - {x}_{2}}$

As the points are $\left(3 , - 2\right)$ and $\left(12 , 19\right)$

the slope of line joining them is (19-(-2))/(12-3 or $\frac{21}{9}$

i.e. $\frac{7}{3}$

Further product of slopes of two lines perpendicular to each other is $- 1$.

Hence slope of line perpendicular to the line passing through (3,−2) and $\left(12 , 19\right)$ will be $- \frac{1}{\frac{7}{3}}$ or $- \frac{3}{7}$.