# How do you write the equation of the line which has y-intercept (0, 5) and is perpendicular to the line with equation y = –3x + 1?

May 29, 2016

$y = \frac{1}{3} x + 5$

#### Explanation:

The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where m represents the gradient and b, the y-intercept.

The advantage of having the line in this form is that m and b can be extracted 'easily'

The equation : y = - 3x + 1 is in this form

hence m = - 3

If 2 lines are perpendicular then the product of their gradients .
${m}_{1} \text{ and " m_2", say,}$ is -1

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{m}_{1.} {m}_{2} = - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

hence gradient of perpendicular line is $\frac{1}{3} \text{ since } \left(3 \times - \frac{1}{3} = - 1\right)$

We have y-intercept = b = 5 and m $= \frac{1}{3}$

$\Rightarrow y = \frac{1}{3} x + 5 \text{ is the equation}$
graph{(y+3x-1)(y-1/3x-5)=0 [-20, 20, -10, 10]}