Question

Find the equation of the straight line that has a $y-$intercept of 5 and is parallel to the straight line $x-3y=3.$

• A. $x-3y+15=0$
• B. $x+3y+5=0$
• C. $x+3y-5=0$
• D. $x-3y-5=0$

Answer 1

The correct option is A $x-3y+15=0$

Given, the equation of the line is $x-3y=3.$
Expressing the lines in $y=mx+c$ format:

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

Comparing this line with $y=mx+c.$
slope of the line $m=\frac{1}{3}$

Slope of the parallel line $m=\frac{1}{3}$
Equation of the line with slope $\frac{1}{3}$and $y-$intercept 5 is
$y=mx+c$
$\Rightarrow y=\frac{1}{3}x+5$
Cross multiply
$\Rightarrow 3y=x+15$
$\Rightarrow x-3y+15=0$