# How do you write an equation of a line passing through (0, 2), perpendicular to y = 4x – 3?

$x + 4 y - 8 = 0$

#### Explanation:

The slope of line: $y = 4 x - 3 \setminus \equiv y = m x + c$ is $4$.

Hence the slope $m$ of the line perpendicular to the given line: $y = 4 x - 3$

$m = - \frac{1}{4}$

Now, the equation of line passing through the point $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(0 , 2\right)$ & having a slope $m = - \frac{1}{4}$ is given as

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

$y - 2 = - \frac{1}{4} \left(x - 0\right)$

$4 y - 8 = - x$

$x + 4 y - 8 = 0$

Jul 21, 2018

$y = - \left(\frac{1}{4}\right) x + 2$

#### Explanation:

First, I'd start with the slope of the new equation. We know that this unknown equation is perpendicular to $y = 4 x - 3$. Since it is perpendicular, the slope of the equation has to be the reciprocal and negative, giving us the new slope of $- \frac{1}{4}$.

So to recap, so far we have $y = - \left(\frac{1}{4}\right) x + b$, with $b$ representing the $y$-intercept.

We know the $y$-intercept is $2$ because $\left(0 , 2\right)$ is a coordinate on the $y$-axis, but I will also solve for it, so it is more clear.

We can find the $y$-intercept by plugging in the coordinates that we are given $\left(0 , 2\right)$ into this new equation and solving for $b$.

$2 = - \left(\frac{1}{4}\right) \left(0\right) + b$

$2 = b$

Now that we have the $y$-intercept, we can put together an equation.
$y = - \left(\frac{1}{4}\right) x + 2$