Question

The triangle formed by the lines $x^2+16xy-11y^2=0,2x+y+1=0$ is

• A. isosceles
• B. right angled
• C. right angled isosceles
• D. equilateral

Answer 1

The correct option is D equilateral

$x^2+16xy-11y^2=0$
Substitute $t=\frac{y}{x}$
$11t^2-16t-1=0$
$\Rightarrow t=\frac{8\pm 5\sqrt{3}}{11}$
Angle between this pair of lines in
$tan\theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$
$=\left|\frac{2\sqrt{(8{)}^2-(-11)}}{(1-11)}\right|$
$=\left|\frac{2\sqrt{75}}{10}\right|$
$\Rightarrow tan\theta =|\sqrt{3}|$
$\Rightarrow \theta ={60}^{\circ }$

$tan\alpha =\frac{8+5\sqrt{3}}{11}$ & $tan\beta =\frac{8-5\sqrt{3}}{11}$

$\therefore tan\theta =\left|\frac{tan\alpha -(-2)}{1-2tan\alpha }\right|=\left|\frac{\frac{8+22-5\sqrt{3}}{11}}{\frac{12-16+10\sqrt{3}}{11}}\right|$

$=\left|\frac{5(6-\sqrt{3})}{5(2\sqrt{3}-1)}\right|$

$=|\sqrt{3}|$

$\therefore \theta ={60}^{\circ }$