# What is the equation of the line perpendicular to y=-7/5  that passes through  (-35,5) ?

Mar 5, 2016

$x = - 35$

#### Explanation:

Firstly, let's go over what we already know from the question. We know that the $y$-$\text{intercept}$ is $- \frac{7}{5}$ and that the slope, or $m$, is $0$.

Our new equation passes through $\left(- 35 , 5\right)$, but the slope will not change since 0 is neither positive nor negative. This means that we need to find the $x - \text{intercept}$. So, our line will be passing through vertically, and have a undefined slope (we don't have to include $m$ in our equation).

In our point, $\left(- 35\right)$ represents our $x - \text{axis}$, and $\left(5\right)$ represents our $y - \text{axis}$. Now, all we have to do is pop the $x - \text{axis}$ $\left(- 35\right)$into our equation, and we're done!

The line that is perpendicular to y=−7/5 that passes through (−35,5) is $x = - 35$.

Here's a graph of both lines.

Mar 5, 2016

solution is, $x + 35 = 0$
$y = - \frac{7}{5}$ represents a straight line parallel to x-axis lying at a distance $- \frac{7}{5}$ unit from x-axis.
Any straight line perpendicular to this line should be parallel to y-axis and can be represented by the equation $x = c$ ,where c = a constant distance of the line from y-axis.
Since the line whose equation to be determined passes through(-35,5) and is parallel to y-axis, it will be at a distance -35 unit from y-axis. Hence its equation should be $x = - 35 \implies x + 35 = 0$