Question

The equation(s) of the straight line passing through $(-2,-7)$ and having intercept length of $3$ units between the straight lines $4x+3y=12$ and $4x+3y=3$ is/are

• A. $7x+24y+182=0$
• B. $x+2=0$.
• C. $y+7=0$
• D. $24x+7y+97=0$

Answer 1

Answer: (A), (B)

The correct options are
A $7x+24y+182=0$
B $x+2=0$.


Given $AB=3$ units
Distance $AC$ between the two given parallel lines
$=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}=\frac{12-3}{\sqrt{16+9}}=\frac{9}{5}$ units
So, $\sin \alpha =\frac{3}{5}$
$\Rightarrow$ $\tan \alpha =\frac{3}{4}$
Let the slope of the required lines be $m$ then
$m=\tan (\theta \pm \alpha )$ where $\tan \theta =-4/3$
$m=\frac{-4/3\pm 3/4}{1\mp 1}=\infty ,-\frac{7}{24}$
So the required line equation's are
$x=-2,7x+24y+182=0$