Question

Vertices of a triangle are $(0,0)$, $(41a,37)$ and $(-37,41b)$ where $a$ and $b$ are the roots of the equation $3x^2-16x+15=0$. The area of the triangle is equal to

Answer 1

Area of triangle = $\frac{1}{2}|0(37-41b)+0(41a+37)+1(41b{)}^2+(37{)}^2|$

$=\frac{1}{2}({41}^{2}ab+{37}^2)=\frac{1}{2}({41}^2\cdot \frac{15}{3}+{37}^2)=4887$

Since a and b are the roots of $3x^2-16x+15=0$, so $ab=\frac{15}{3}$