# Straight lines, Combined equation.

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.

• Factorise $(a^2-3b^2)x^2+8abxy+(b^2-3a^2)y^2=0$ Commented Aug 9, 2016 at 14:29
• ok..then what do i get?Please, explain the whole solution cause this question is totally new to me. Commented Aug 9, 2016 at 14:30
• Factoring the first equation gives you the individual equations of the two lines. From there you can proceed in various ways, such as finding their points of intersection and showing that those form an equilateral triangle. Alternatively, you could compare the angles that the three lines make with each other, and then use the distance of the third line from the origin to compute the triangle’s area.
– amd
Commented Aug 9, 2016 at 19:46

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.

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*}

• @muralidharanan I didnt know about the equation of the angle bisectors of a pair of straight lines....Btw.. what's the proof for the formula? Commented Aug 12, 2016 at 12:12
• Let the angles made by the lines $ax^2+2hxy+by^2=0$ with the $X-$axis be $\theta_1$ and $\theta_2$. We have \begin{align*} \tan\theta_1 + \tan\theta_2 &= -\frac{2h}{b} \\ \tan\theta_1\tan\theta_2 &= \frac{a}{b} \end{align*} Then the angles made by the bisectors of the angles are $\theta = \frac{\theta_1+\theta_2}{2}$ and $90^{\circ} + \theta$. Thus if the slopes of the bisectors are $m_3$ and $m_4$, then \begin{align*} m_3 + m_4 &= \frac{b-a}{h} \end{align*} and $m_3m_4=-1$. Use these two get the equation of the bisectors.
– user348749
Commented Aug 12, 2016 at 15:42