Question

The lines that are parallel to $2x+6y-7=0$ and have an intercept $10$ units between coordinate axes is/are

• A. $y=\frac{-1}{3}x+\sqrt{10}$
• B. $y=\frac{1}{3}x+\sqrt{10}$
• C. $y=\frac{-1}{3}x-\sqrt{10}$
• D. $y=\frac{1}{3}x-\sqrt{10}$

Answer 1

Answer: (A), (C)

$y=\frac{-1}{3}x-\sqrt{10}$

The slope of the given line is $m=\frac{-1}{3}$ Therefore, the slope of any line parallel to the given line is also $m=\frac{-1}{3}$
The slope intercept form of any line parallel to the given line is $y=$ $\frac{-1}{3}x+c$ where $c$ is the $y$ -intercept.
Thus, the $x$ -intercept is $(3c,0)$ and the $y$ -intercept is $(0,c)$
The distance between the intercepts is to be $10$ units, therefore, using the distance formula
$10=\sqrt{(3c{)}^2+c^2}$
$100=9c^2+c^2$
$c=\pm \sqrt{10}$
There are two lines parallel to the given line, such that the distance between the $x$ and $y$ intercepts is $10$ units, their equations are:
$y=\frac{-1}{3}x+\sqrt{10}$
and
$y=\frac{-1}{3}x-\sqrt{10}$