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Line in second quadrant

Joe_1234

New member
May 15, 2020
25
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

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,666
St. Augustine, FL.
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

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,666
St. Augustine, FL.
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

New member
May 15, 2020
25
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