Show that the lines $x^2+16xy-11y^2=0$ form an equilateral triangle with the line $2x+y+1=0$ and find its area.
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My solution is here; Here $x^2+16xy-11y^2=0$ represents a pair of straight lines. Comparing with $ax^2+bx+c=0$ and using the quadratic formula i.e. $x={-b+/-√(b^2-4ac)}/2a$ I got $x=-8y+/-5y√3$. Now, how do i move ahead
The pair of lines intersect at the origin, so let each line be $y=mx$. Substitute for $y$: $$x^2+16mx^2-11m^2x^2=0$$
This gives the quadratic equation, satisfied by both gradients:
$$11m^2-16m-1=0\Rightarrow m=\frac{8\pm5\sqrt{3}}{11}$$
Now consider the line through the origin which is perpendicular to the third side $2x+y+1=0$ which is $$2y=x$$
We can rotate this line by $\pm 30^o$ by using the rotation matrix $$Rot(\alpha)=\left(\begin{matrix}\cos \alpha &-\sin \alpha \\ \sin\alpha& \cos\alpha\end{matrix}\right)$$
This leads to two lines, passing through the origin, whose gradients are exactly as before, $\frac{8\pm5\sqrt{3}}{11}$.
This confirms that the triangle formed by these lines is equilateral.
To calculate the area, you just need the height of the triangle, which is the perpendicular distance from the origin to the line $2x+y+1$, namely $$\frac{1}{\sqrt{5}}$$ then the area of the triangle is $$\frac 15 \tan 30$$.
I hope this helps.