For line $ax+by+k=0$ which intercepts form a triangle rectangle with area $A$, find $k$

Question

I know that the area of a triangle is given by the formula $A=\frac{1}2Bh$ and the intercepts of line $ax+bx+k=0$ are $(B,0)$ and $(0,h)$ which forms a square with area $2A$, but without brute-forcing my way through it, I got no idea how can I find this the analytic way. For example, for line $4x-5y+k=0$ and area $A=2\frac{1}2$, $k$ is $\pm10$, but how I know is?

Answer 1

To find the intercepts, plug in $x=0$ and $y=0$ to $ax + by + k = 0$. We see that it has intercepts of $(0, -k/b)$ and $(-k/a, 0)$.

Therefore, the area of this triangle along with $(0,0)$ is $$A = \dfrac{1}{2} \cdot Bh = \dfrac{k^2}{2|ab|}$$.

Therefore, when $a=4, b=5, A = 2.5$, we have $$2.5 =\dfrac{k^2}{2 \cdot 4 \cdot 5} \implies k^2 = 100 \implies k=10,-10$$