# How do you write an equation of a line passing through (2, 5), perpendicular to x - 5y = -10?

$\left(y - 5\right) = - 5 \left(x - 2\right) \iff y = - 5 x + 15$

#### Explanation:

Let's first look at the the line $x - 5 y = - 10$ and solve it for $y$ so that we can put it into the slope-intercept form, the general form of which is:

$y = m x + b$, where $m =$ slope and $b =$ y-intercept.

$x - 5 y = - 10$

$5 y = x + 10$

$y = \frac{1}{5} x + 2$

And so slope, $m = \frac{1}{5}$

To find the perpendicular slope, we take the negative inverse, which gives us $m = - 5$.

To find the line passing through the point $\left(2 , 5\right)$ with slope $- 5$, we can use the point slope form, which has as a general form:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$ where ${x}_{1} , {y}_{1}$ are a point. So we have:

$\left(y - 5\right) = - 5 \left(x - 2\right)$

For comparison, let's put this into slope-intercept form:

$y = - 5 x + 15$

And we can graph them both:

graph{(y-(-5x+15))(y-(1/5x+2))=0 [-10, 10, -8, 8]}