# What is the equation of the line that passes through (0,-1) and is perpendicular to the line that passes through the following points: (13,20),(16,1) ?

##### 1 Answer
Feb 13, 2016

$y = \frac{3}{19} \cdot x - 1$

#### Explanation:

The slope of the line passes through (13,20) and (16,1) is ${m}_{1} = \frac{1 - 20}{16 - 13} = - \frac{19}{3}$ We know condition of perpedicularity between two lines is product of their slopes equal to be -1 $\therefore {m}_{1} \cdot {m}_{2} = - 1 \mathmr{and} \left(- \frac{19}{3}\right) \cdot {m}_{2} = - 1 \mathmr{and} {m}_{2} = \frac{3}{19}$ So the line passing through (0,-1) is $y + 1 = \frac{3}{19} \cdot \left(x - 0\right) \mathmr{and} y = \frac{3}{19} \cdot x - 1$graph{3/19*x-1 [-10, 10, -5, 5]} [Ans]