Linear equation and triangles

Question

Given a linear equation pass through the point(6,4) and two axes, and formed a triangle with area 6. I want to find the equation of that line.

My attempt is letting the equation be $\frac{y-4}{x-6}=m$ Then using the fact that the area of triangle is 6, forming the equation $\frac{1}{2}\text{base}*\text{height}=6$. I got the base and height of the triangle from the axes, $\text{base} = 6-\frac{4}{m}$ and $\text{height}=4-6m$

Then substitute the equation $\frac{1}{2}\text{base}*\text{height}=6$ and get $36m^2-36m+16=0$ which leads to no solution. Can anyone let me know what's wrong?

Answer 1

You get if calculated correctly $36m^2-60m+16=0$

Answer 2

The equation is given by $$y=mx+n$$ so $$y=m(x-6)+n$$ since $$P(6;4)$$ is situated on the line. An both intersection Points are $$A(0;4-6m)$$ and $$P(\frac{6m-4}{m},0)$$. The area our triangle is giveb by$$A=\frac{1}{2}\left|4-6m\right|\left|\frac{6m-4}{m}\right|$$