# What is the equation of the line that passes through (6,-1) and is perpendicular to the line that passes through the following points: (8,-3),(12,10) ?

##### 1 Answer
Mar 18, 2016

$y = - \frac{4}{13} x + \frac{11}{13}$

#### Explanation:

${P}_{1} \left(6 , - 1\right)$
${P}_{A} \left(x , y\right) \text{ any point on line pases trough (6,-1)}$
${m}_{1} = \frac{y - \left(- 1\right)}{x - 6}$
${m}_{1} = \frac{y + 1}{x - 6} \text{ slope of line}$

${m}_{2} = \frac{10 - \left(- 3\right)}{12 - 8}$
${m}_{2} = \frac{13}{4} \text{ slope of other line pases trough (8,-3)(12,10)}$
${m}_{1} \cdot {m}_{2} = - 1 \text{ (if lines are perpendicular)}$

$\frac{y + 1}{x - 6} \cdot \frac{13}{4} = - 1$

$\frac{13 y + 13}{4 x - 24} = - 1$

$13 y + 13 = - 4 x + 24$

$13 y = - 4 x + 24 - 13$

$13 y = - 4 x + 11$

$y = - \frac{4}{13} x + \frac{11}{13}$