Question

How do you write an equation of a line given $(-3,6)$and has slope $\frac{1}{3}$?

Answer 1

Find the equation of a line:

The equation of the line is obtained by the formula $y-y_1=m\left(x-x_1\right)$, where $m$ is the slope and $\left(x_1,y_1\right)$ is a point on the line.

It is given that point $(-3,6)$lies on the lien and the slope of the equation is $\frac{1}{3}$.

Put $x_1=-3$,$y_1=6$ and $m=\frac{1}{3}$ in $y-y_1=m\left(x-x_1\right)$.

$\begin{array}{rcl}y-6& =& \frac{1}{3}\left(x-\left(-3\right)\right)\\ & \Rightarrow & y-6=\frac{1}{3}\left(x+3\right)\\ & \Rightarrow & 3\left(y-6\right)=x+3\left[\mathrm{Multiply}\mathrm{both}\mathrm{sides}\mathrm{by}3\right]\\ & \Rightarrow & 3y-18=x+3\\ & \Rightarrow & 3y-18+18-x=x+3+18-x\left[\mathrm{Add}\left(18-\mathrm{x}\right)\mathrm{from}\mathrm{both}\mathrm{sides}\right]\\ \therefore 3y-x& =& 21\end{array}$

Hence, the equation of the line is $3y-x=21$.