Question

Find the equation of the line such that the area of the triangle formed by the line and the coordinate axes in the first quadrant is $30$ and length of the hypotenuse is $13$.

Answer 1

$\frac{1}{2}ab=30$

$ab=60$ $a^2+b^2=169$

$(a+b{)}^2-2ab=169$

$(a+b{)}^2-120=169$

$(a+b{)}^2=289=(17{)}^2$

$a+b=17$

$a+\frac{60}{a}=17$

$a^2-17a+60=0$

$a^2-12a-5a+60=0$

$a(a-12)-5(a-12)=0$

$a=5,a=12$

$a=5,b=12$ or $a=12,b=5$

equation $\to \frac{x}{3}+\frac{y}{12}=1$ or $\frac{x}{12}+\frac{y}{5}=1$d