Question

The equation of a straight line(s) passing through $(1,2)$ and having intercept of length $3$ units between the straight lines $3x+4y=24$ and $3x+4y=12$, is

• A. $y=2$
• B. $x=1$
• C. $7x+24y+41=0$
• D. $7x-24y+41=0$

Answer 1

Answer: (B), (D)

$7x-24y+41=0$

​​​ Let the line passing through $(1,2)$ makes an angle of $\alpha$ with the given parallel lines


Now the distance between the parallel lines is $\frac{12}{5}$ units
So, from diagram $\tan \alpha =\frac{4}{3}$
If the slope of the required line is $m$, then $m=$ tan$(\theta \pm \alpha )$ , where tan$\theta$ is slope of given line = $-\frac{3}{4}$
$m=\frac{-3/4\pm 4/3}{1\mp 1}=\infty ,\frac{7}{24}$
So the equation of required lines are
$x=1,7x-24y+41=0$