# How do you write the equation that represents the line perpendicular to y=-3x+ 4 and passing through the point (-1, 1)?

Dec 13, 2016

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

#### Explanation:

First, we need to determine the slope of the perpendicular line.

If the slope of a line is $\frac{a}{b}$ then the slope of a line perpendicular to this is $- \frac{b}{a}$.

Because the equation for the line we are given is already in slope-intercept form we can obtain the slope.

Slope-intercept form is $y = m x + c$ where $m$ is slope.

So, the slope of the line given is:

$- 3$ or $- \frac{3}{1}$ converting this to the slope of a perpendicular line using the rule above gives:

$- - \frac{1}{3} \to \frac{1}{3}$

Now that we have the slope and a point we can use the point-slope formula to determine the equation for the perpendicular line.

The point-slope formula is:

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

Where $m$ is the slope and

$\left({x}_{1} , y 1\right)$ is the point which is given. Substituting the information gives:

$y - 1 = \frac{1}{3} \left(x - - 1\right)$

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

To convert to slope-intercept form we can solve for $y$ while keeping the equation balanced:

$y - 1 = \frac{1}{3} x + \frac{1}{3} \times 1$

$y - 1 = \frac{1}{3} x + \frac{1}{3}$

$y - 1 + 1 = \frac{1}{3} x + \frac{1}{3} + 1$

$y - 0 = \frac{1}{3} x + \frac{1}{3} + \frac{3}{3}$

$y = \frac{1}{3} x + \frac{4}{3}$