Question

Find the equation of a line passing through the point $(2,3)$ and parallel to the line $2x-3y+8=0$.

• A. $2x-3y=-5$
• B. $2x-3y=12$
• C. $x-3y=4$
• D. $3x-2y=7$

Answer 1

Answer: (A)

$2x-3y=-5$

First we find the slope of the line $2x-3y+8=0$ by placing it into slope intercept form:

$2x-3y+8=0$

$\Rightarrow -3y=-2x-8$

$\Rightarrow 3y=2x+8$

$\Rightarrow y=\frac{2}{3}x+\frac{8}{3}$

Therefore, the slope of the line is $m=\frac{2}{3}$.

Now since the equation of the line with slope $m$ passing through a point $(x_1,y_1)$ is

$y-y_1=m(x-x_1)$

Here the point is $(2,3)$ and slope is $m=\frac{2}{3}$, therefore, the equation of the line is:

$y-3=\frac{2}{3}(x-2)\Rightarrow 3(y-3)=2(x-2)\Rightarrow 3y-9=2x-4\Rightarrow 2x-3y=-9+4\Rightarrow 2x-3y=-5$

Hence, the equation of the line is $2x-3y=-5$.