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#### Joe_1234

##### New member
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.

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#### MarkFL

Staff member
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

Staff member

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

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

##### New member

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

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