Question

Find the equation of the line which has positive y-intercept $4$ units and is parallel to the line $2x-3y-7=0$. Also find the point where it cuts the x-axis.

• A. $2x-3y+12=0,(6,0)$
• B. $2x-3y+12=0,(-6,0)$
• C. $2x-3y+4=0,(2,0)$
• D. $2x+3y+4=0,(-2,0)$

Answer 1

Answer: (B)

$2x-3y+12=0,(-6,0)$

$2x-3y-7=0\Rightarrow 3y=2x-7$

$\Rightarrow y=\frac{2}{3}x-\frac{7}{3}$

Equation of line parallel to this is

$y=\frac{2}{3}x+c$

This line make y intercept $=4$ hence

$c=4$

Therefore equation of line is

$y=\frac{2}{3}x+4\Rightarrow 2x-3y+12=0$

And point is $(-6,0).$