# How do you find the line that is perpendicular to y=2/3x+1 and passes through (0, 5)?

Nov 10, 2016

$y = - \frac{3}{2} x + 5$

#### Explanation:

Perpendicular lines have opposite reciprocal slope, $m$, in an equation. Meaning, if the slope of a line is a positive fraction, its reciprocal slope would be negative.

The given equation is a linear equation, $y = m x + b$

Where $m$ is the slope, (rise)/(run, and $b$ is the y intercept

To find a line that is perpendicular to the given equation, use the opposite reciprocal slope and point and solve using the point-slope formula

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

$m = - \frac{3}{2}$

${y}_{1} = 5$

${x}_{1} = 0$

$y - 5 = - \frac{3}{2} \left(x - 0\right)$

Distribute the $- \frac{3}{2}$ throughout the set of parenthesis

$y - 5 = - \frac{3}{2} x + 0$

Add $5$ on both sides of the equation

$y = - \frac{3}{2} x + 5$

And now you can see the slope is the opposite reciprocal of the original equation

Nov 10, 2016

$y = - \frac{3}{2} x + 5$

#### Explanation:

If lines are perpendicular, one slope is the negative reciprocal of the other.
(Their product is -1)

If ${m}_{1} = \frac{2}{3} , \text{ } {m}_{2} = - \frac{3}{2}$

So we have the slope of the line perpendicular to the one given.

The point $\left(0 , 5\right)$ is the y-intercept because the x-value is 0.
This is also known as 'c' $\rightarrow c = 5$

The equation of a line is $y = m x + c$.
We have $m \mathmr{and} c$

$y = - \frac{3}{2} x + 5$