# The equation of line QR is y = - 1 /2 x + 1. How do you write an equation of a line perpendicular to line QR in slope-intercept form that contains point (5, 6)?

Aug 25, 2017

See a solution process below:

#### Explanation:

First, we need to find the slope of the for the two points in the problem. The line QR is in slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y = \textcolor{red}{- \frac{1}{2}} x + \textcolor{b l u e}{1}$

Therefore the slope of QR is: $\textcolor{red}{m = - \frac{1}{2}}$

Next, let's call the slope for the line perpendicular to this ${m}_{p}$

The rule of perpendicular slopes is: ${m}_{p} = - \frac{1}{m}$

Substituting the slope we calculated gives:

${m}_{p} = \frac{- 1}{- \frac{1}{2}} = 2$

We can now use the slope-intercept formula. Again, the slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

Substituting the slope we calculated gives:

$y = \textcolor{red}{2} x + \textcolor{b l u e}{b}$

We can now substitute the values from the point in the problem for $x$ and $y$ and solve for $\textcolor{b l u e}{b}$

$6 = \left(\textcolor{red}{2} \times 5\right) + \textcolor{b l u e}{b}$

$6 = 10 + \textcolor{b l u e}{b}$

$- \textcolor{red}{10} + 6 = - \textcolor{red}{10} + 10 + \textcolor{b l u e}{b}$

$- 4 = 0 + \textcolor{b l u e}{b}$

$- 4 = \textcolor{b l u e}{b}$

Substituting this into the formula with the slope gives:

$y = \textcolor{red}{2} x + \textcolor{b l u e}{- 4}$

$y = \textcolor{red}{2} x - \textcolor{b l u e}{4}$