The area (in square units) of the triangle formed by the lines $6 x^2+13 x y+6 y^2=0$ and $x+2 y+3=0$ is
The area (in square units) of the triangle formed by the lines $6 x^2+13 x y+6 y^2=0$ and $x+2 y+3=0$ is
$\frac{9}{2}$
$\frac{45}{4}$
$\frac{9}{8}$
$\frac{45}{8}$
Solution
Given the curve, $6 x^2+13 x y+6 y^2=0$
Comparing with $a x^2+2 h x y+b y^2=0$,
we get $a=6, b=6, h=\frac{13}{2}$
and a line $x+2 y+3=0$ comparing with $l x+m y+n=0$,
we get, $l=1, m=2, n=3$
So, required area $=\frac{n^2 \sqrt{h^2-a b}}{\left|a m^2-2 h l m+b l^2\right|}$
$=\frac{9 \sqrt{\frac{169}{4}-36}}{|6 \times 4-13 \times 1 \times 2+6 \times 1|}=\frac{9 \times \frac{5}{2}}{4}=\frac{45}{8}$ sq. units.