Equation of Line in Second Quadrant with Area 4 and Differing Intercepts

[Joe_1234]

Find the equation of the line which forms with the axes in the second quadrant a triangle of area 4 and whose intercepts differ by 5.

 

[MarkFL]

Find the equation of the line which forms with the axes in the second quadrant a triangle of area 4 and whose intercepts differ by 5.

I would begin with the two-intercept equation of a line:

\(\displaystyle \frac{x}{a}+\frac{y}{b}=1\)

Where:

\(\displaystyle (a-b)^2=5^2\)

\(\displaystyle \frac{1}{2}(-a)b=4\)

We have two equations in two unknowns...can you proceed (observing that \(a<0<b\))?

 

[MarkFL]

Let's follow up...we have:

\(\displaystyle ab=-8\implies b=-\frac{8}{a}\)

And so:

\(\displaystyle \left(a+\frac{8}{a}\right)^2=5^2\)

\(\displaystyle \frac{a^2+8}{a}=\pm5\)

\(\displaystyle a^2\pm5a+8=0\)

\(\displaystyle a=\frac{\pm5\pm\sqrt{5^2-32}}{2}\)

And since the discriminant is negative, we find there is no real solution.

 

[Joe_1234]

Let's follow up...we have:

\(\displaystyle ab=-8\implies b=-\frac{8}{a}\)

And so:

\(\displaystyle \left(a+\frac{8}{a}\right)^2=5^2\)

\(\displaystyle \frac{a^2+8}{a}=\pm5\)

\(\displaystyle a^2\pm5a+8=0\)

\(\displaystyle a=\frac{\pm5\pm\sqrt{5^2-32}}{2}\)

And since the discriminant is negative, we find there is no real solution.

Thank you sir