Straight lines, Combined equation.

Question

Show that the straight lines $(A^2 - 3B^2)x^2 + 8ABxy + (B^2-3A^2)y^2=0$ form with the line $Ax+By+C=0$ is an equilateral triangle of area $C^2/[\sqrt3(A^2 +B^2)]$.
I have no idea how to start\attempt this question. Please, help.

Answer 1

The angle $\theta$ between the lines $ax^2+2hxy+by^2 = 0$ is given by \begin{align*} \tan\theta = \frac{2\sqrt{h^2-ab}}{a+b} \end{align*} In this case, we have \begin{align*} \tan\theta &= \pm\frac{2\sqrt{16A^2B^2-(A^2-3B^2)(B^2-3A^2)}}{-2(B^2+A^2)}\\ &=\pm \sqrt{3} \end{align*} Thus the angles between the lines is $60^\circ$. Also, the angle bisectors of the lines $ax^2+2hxy+by^2 = 0$ are given by \begin{align*} \frac{x^2-y^2}{a-b} = \frac{xy}{h} \end{align*} Here the angle bisectors are given by \begin{align*} \frac{x^2-y^2}{A^2-B^2} = \frac{xy}{AB} \end{align*} This can be written as \begin{align*} (Ax+By)(Bx-Ay) = 0 \end{align*} Note that $Ax+By=0$ is parallel to $Ax+By+C=0$ and $Bx-Ay=0$ is perpendicular to it. Thus the angle bisector is perpendicular to the third side and the triangle is isosceles. Since the vertex angle is $60^\circ$, it follows that the triangle is equilateral.

We just need to find the altitude of the triangle to compute the area. This is equal to the perpendicular distance from the origin on $Ax+By+C = 0$ and equals \begin{align*} \frac{C}{\sqrt{A^2+B^2}} \end{align*} Since the area is $\frac{h^2}{\sqrt{3}}$, it follows that the area is \begin{align*} \frac{C^2}{\sqrt{3}(A^2+B^2)} \end{align*}

Answer 2

1. Apply factorization to decompose the line pair into:- $$ L_1: (A + \sqrt B)x + (B - \sqrt A)y = 0$$ and $$L_2 : (A - \sqrt B)x + (B + \sqrt A)y = 0$$.

2. Find $m(L_1)$ and $m(L_2)$; where $m(L_1) = - \dfrac {(A + \sqrt B)}{ (B - \sqrt A)}$ and $m(L_2) = …$.

3. Apply the formula that can find $\theta$, the angle between $L_1$ and $L_2$ using $m(L_1)$ and $m(L_2)$. Should be able to get $\tan \theta = …. = \sqrt 3$ indicating one of the angles formed is $60^0$.

4. Let $L_3 : Ax + By + C = 0$. Then $m(L_3) = \dfrac {-A}{B}$.

5. Repeat step 3 to show the angle between $L_1$ and $L_3$ is also $60^0$. The process of proving the triangle formed is equilateral is then complete.

The next part is to prove the area of the triangle thus formed is equal to the given value.

Noting that $L_1$ and $L_2$ intersect at O(0, 0), we can use an alternate (easier) way by finding h instead; where h is the normal distance of $L_3$ from O. Actually, $h = … = | (\dfrac {C}{\sqrt (A^2 + B^2)}|$ is the altitude of the said triangle.

If h is known, the corresponding area can be found because the object is an equilateral triangle.