# How do you write an equation in point slope and slope intercept form given (8, -2) and perpendicular to the line whose equation is x-5y-7=0?

Jun 5, 2018

Point slope form equation: $\left(y + 2\right) = - 5 \left(x - 8\right)$
Slope intercept form equation: $y = - 5 x + 38$

#### Explanation:

Slope of the line,  x-5 y -7=0 or y= 1/5 x -7/5; [y=m x+c]

is ${m}_{1} = \frac{1}{5}$ [Compared with slope-intercept form of equation]

The product of slopes of the pependicular lines is ${m}_{1} \cdot {m}_{2} = - 1$

$\therefore {m}_{2} = \frac{- 1}{\frac{1}{5}} = - 5$. In point slope form, equation of line

passing through $\left({x}_{1} = 8 , {y}_{1} = - 2\right)$ having slope of ${m}_{2} = - 5$

is $y - {y}_{1} = {m}_{2} \left(x - {x}_{1}\right) \therefore y - \left(- 2\right) = - 5 \left(x - 8\right)$or

$\left(y + 2\right) = - 5 \left(x - 8\right)$. Slope intercept form:

$\left(y + 2\right) = - 5 \left(x - 8\right) \mathmr{and} y + 2 = - 5 x + 40$ or

$y = - 5 x + 40 - 2 \mathmr{and} y = - 5 x + 38$ [Ans]