Question

The equation of a straight line, perpendicular to $3x-4y=6$ and forming a triangle of area $6$ square units with coordinate axes, is

• A. $x-2y=6$
• B. $4x+3y=12$
• C. $4x+3y+24=0$
• D. $3x+4y=12$

Answer 1

Answer: (B)

$4x+3y=12$

The line which is perpendicular to the given line will have a slope of $\frac{-4}{3}$

The equation of the line will look like $3y+4x=k$

This line intersects the $x$-axis at $(\frac{k}{4},0)$ an $y$-axis at $(0,\frac{k}{3}).$

The area of the triangle formed is given by $\frac{1}{2}\times \frac{k}{3}\times \frac{k}{4}=6$

$\Rightarrow k^2=144$

$\Rightarrow k=\pm 12$

Thus, the line equation becomes $4x+3y=\pm 12$