# What is the equation of the line that passes through (0,-1) and is perpendicular to the line that passes through the following points: (8,-3),(1,0) ?

Feb 10, 2016

$7 x - 3 y + 1 = 0$

#### Explanation:

Slope of the line joining two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by

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

As the points are $\left(8 , - 3\right)$ and $\left(1 , 0\right)$, slope of line joining them will be given by $\frac{0 - \left(- 3\right)}{1 - 8}$ or $\frac{3}{- 7}$

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

Product of slope of two perpendicular lines is always $- 1$. Hence slope of line perpendicular to it will be $\frac{7}{3}$ and hence equation in slope form can be written as

$y = \frac{7}{3} x + c$

As this passes through point $\left(0 , - 1\right)$, putting these values in above equation, we get

$- 1 = \frac{7}{3} \cdot 0 + c$ or $c = 1$

Hence, desired equation will be

$y = \frac{7}{3} x + 1$, simplifying which gives the answer

$7 x - 3 y + 1 = 0$