What is the equation of the line perpendicular to y=-3/4x  that passes through  (2,4) ?

Aug 23, 2017

$y = \frac{4}{3} x + \frac{4}{3}$

Explanation:

We begin by finding the slope of the line that is perpendicular to $- \frac{3}{4}$. Recall that the perpendicular slope is expressed as the negative reciprocal of the slope($m$) or $- \frac{1}{m}$.

Therefore, if the slope is $- \frac{3}{4}$ the perpendicular slope is...

$- \frac{1}{- \frac{3}{4}} \to - 1 \cdot - \frac{4}{3} = \frac{4}{3}$

Now that we have the perpendicular slope, we can find the equation of the line by using the point-slope formula: $y - {y}_{1} = m \left(x - {x}_{1}\right)$ where $m$ is the slope and $\left(2 , 4\right) \to \left({x}_{1} , {y}_{1}\right)$

So to find the equation of the line...

$y - 4 = \frac{4}{3} \left(x - 2\right) \leftarrow$ Equation of the line

We can also rewrite the above equation in $y = m x + b$ form if desired. To do this, we simply solve for $y$:

$y - 4 = \frac{4}{3} x - \frac{8}{3}$

$y - 4 = \frac{4}{3} x - \frac{8}{3}$

$y \cancel{- 4} \cancel{\textcolor{red}{+ 4}} = \frac{4}{3} x - \frac{8}{3} \textcolor{red}{+ 4}$

$y = \frac{4}{3} x - \frac{8}{3} + \left[\frac{4}{1} \left(\frac{3}{3}\right)\right]$

$y = \frac{4}{3} x - \frac{8}{3} + \frac{12}{3}$

$y = \frac{4}{3} x + \frac{4}{3}$