The discussion focuses on finding the equation of a line in the second quadrant that forms a triangle with the axes, having an area of 4 and intercepts differing by 5. The two-intercept form of the line is represented as $$\frac{x}{a}+\frac{y}{b}=1$$, leading to the equations $$(a-b)^2=25$$ and $$\frac{1}{2}(-a)b=4$$. Upon solving, it is determined that the discriminant of the resulting quadratic equation is negative, indicating that there is no real solution for the given conditions.
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Joe_1234 said: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.
Thank you sirMarkFL said:Let's follow up...we have:
$$ab=-8\implies b=-\frac{8}{a}$$
And so:
$$\left(a+\frac{8}{a}\right)^2=5^2$$
$$\frac{a^2+8}{a}=\pm5$$
$$a^2\pm5a+8=0$$
$$a=\frac{\pm5\pm\sqrt{5^2-32}}{2}$$
And since the discriminant is negative, we find there is no real solution.