# What is the equation of the line perpendicular to y=-15/7x  that passes through  (-1,7) ?

Jul 1, 2018

Point-slope form: $y - 7 = \frac{7}{15} \left(x + 1\right)$

Slope-intercept form: $y = \frac{7}{15} x + \frac{112}{15}$

#### Explanation:

The slope of a perpendicular line is the negative reciprocal of the original slope. In this case, the perpendicular slope of $- \frac{15}{7}$ is $\frac{7}{15}$. The product of two perpendicular slopes is $- 1$.

$- \frac{15}{7} \times \frac{7}{15} = - 1$

With the slope and one point, you can write a linear equation in point-slope form:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$,

where:

$m$ is the slope, and $\left({x}_{1} , {y}_{1}\right)$ is the given point.

Plug in the known values.

$y - 7 = \frac{7}{15} \left(x - \left(- 1\right)\right)$

Simplify.

$y - 7 = \frac{7}{15} \left(x + 1\right)$

You can convert the point-slope form to slope-intercept form by solving for $y$. $\left(y = m x + b\right)$

$y = \frac{7}{15} x + \frac{7}{15} + 7$

Multiply $7$ by $\frac{15}{15}$ to get an equivalent fraction with the denominator $15$.

$y = \frac{7}{15} x + \frac{7}{15} + 7 \times \frac{15}{15}$

$y = \frac{7}{15} x + \frac{7}{15} + \frac{105}{15}$

$y = \frac{7}{15} x + \frac{112}{15}$ $\leftarrow$ slope-intercept form

graph{y-7=7/15(x+1) [-10.04, 9.96, 1.44, 11.44]}