Finding the equation of a line given a line perpendicular to it and the area of a triangle with the axes

Question

A line $l$ is perpendicular to the line $3x-4y+18=0$ and the area of triangle bounded by the line $l$ with the co ordinate axes is $6$ sq. units, find the equation of $l$.

My Approach,

Since the line $l$ is perpendicular to the line $3x-4y+18=0$, its equation must be $$4x+3y+k=0$$. But I don't know the point through which this equation passes. Then, how do I find the value of $k$.

I got stuck at here. Please help me to complete.

Answer 1

The triangle of area $6$ is determined by three points:

• The origin of coordinates, $(0,0),$
• The cut point of the line $l$ and the $x$-axis, and
• The cut point of the line $l$ and the $y$-axis.

Since the equation of $x$-axis is $y=0$ we solve

$$\left\{\begin{align}y & =0 \\ 4x+3y+k &=0\end{align}\right.$$ to get the cut point $(-k/3,0).$

In a similar way we get the cut point of $l$ and the $y$-axis, getting the point $(0,-k/3).$

So, the area of the right triangle is $$\frac{1}{2}\cdot \frac{k}{3}\cdot \frac{k}{4}.$$ If we equal this quantity to $6$ we get that $k=\pm 12.$

Answer 2

Here is the picture drawn out as requested apart from the well written answer above.