Question

Given lines : $4x+3y=3$ and $4x+3y=12$. One of the possible equations of straight line passing through $(-2,-7)$ and making an intercept of length $3$ between the given lines is

• A. $y=-7$
• B. $4x+3y=-29$
• C. $3x-4y=22$
• D. $7x+24y+182=0$

Answer 1

Answer: (D)

$7x+24y+182=0$

Distance between two parallel lines
$d=\frac{|-9|}{\sqrt{9+16}}=\frac{9}{5}$
So, $AC=\frac{9}{5}$
$\Rightarrow BC=\frac{12}{5}$
Let $\theta$ be the angle between AB and BC
$\Rightarrow \tan \theta =\frac{3}{4}$
Also, $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
where $m_1$ is slope of required line $AB$ and $m_2$ is the slope of line BC
$\Rightarrow \pm \frac{3}{4}=\frac{m_1+\frac{4}{3}}{1-\frac{4}{3}{m}_1}$
$\Rightarrow m_1=-\frac{7}{24}$
Hence, the equations of AB are
$y+7=-\frac{7}{24}(x+2)$
$\Rightarrow 7x+24y+182=0$