If lines $x+y=7$ and $ax^2+2hxy+ay^2=0$ ($a\neq 0$) form a triangle, then show that the triangle is isosceles 
If the lines $x+y=7$ and $ax^2+2hxy+ay^2=0$ (with $a\neq 0$) form a triangle, then what kind of triangle do they form?
Answer: isosceles

I myself tried to find the slopes and compare the angles, but wasn't able to solve it.
 A: The symmetry of an isosceles triangle is better viewed with a rotation of the coordinates by 45 degrees, i.e. with the coordinate change below,
$$ x=u+v, \>\>\>\>\>y=u-v$$
Then, the quadratic equation becomes $(h+a)u^2-(h-a)v^2=0$, or,
$$\left(u+\sqrt{\frac{h-a}{h+a}}v\right)\left(u-\sqrt{\frac{h-a}{h+a}}v\right)=0\tag{1}$$
and $x+y=7$ simplifies to,
$$u=\frac 72$$
Now, we can conveniently observe that the line $u=\frac 72$ is the base of the triangle, and the pair of lines from (1),
$$u=\sqrt{\frac{h-a}{h+a}}v,\>\>\>\>\>u=-\sqrt{\frac{h-a}{h+a}}v$$
are the two isosceles sides due to their opposite slopes. It is an upside down isosceles triangle in the $uv$-coordinates. 
A: Any point on $x+y=7$ can be chosen to be $P(t,7-t)$
Let us find the intersection,
$$0=at^2+2ht(7-t)+a(7-t)^2=t^2(2a-2h)+2t(7h-7a)+49a$$
$$t_1+t_2=\dfrac{14(a-h)}{2(a-h)}=7$$ assuming $a-h\ne0$ as $a=h$ will make both lines identical
The midpoint$(M)$ of $x+y=7$ $$\left(\dfrac72,\dfrac{14-(t_1+t_2)}2\right)\equiv\left(\dfrac72,\dfrac72\right)$$
The intersection of the two lines $$ax^2+2hxy+ay^2=0$$ is $O(0,0)$
So, the gradient of $OM$ will be $$=\dfrac{\dfrac72}{\dfrac72}=1$$ which is perpendicular to $x+y=7$
So, the triangle is at least isosceles. See this
A: Hint:
Let $\varphi$ the polar angle of the line with equation $x+y=7$, $\theta, \theta'$ the polar angles of the lines defined by the equation $x^2+2hxy+ay^2=0\iff x^2+2\frac ha xy+y^2=0$.
We may conjecture that the basis of the isosceles triangle is the side on the first  line, so we only have to show that $|\theta-\varphi|=|\theta'-\varphi|$, which is equivalent to
$$\bigl|\tan(\theta-\varphi)\bigr|=\bigl|\tan(\theta'-\varphi)\bigr|\iff \biggl|\frac{\tan\theta-\tan\varphi}{1+\tan\theta\tan\varphi}\biggr|= \biggl|\frac{\tan\theta'-\tan\varphi}{1+\tan\theta'\tan\varphi}\biggr|$$
Now  $\tan\varphi=-1$ and $t=\tan\theta$, $t'=\tan\theta'$ are the roots of the quadratic equation $\;t^2+2\frac ha t+1=0$, so $tt'=1$.
Some details:
We have to prove that
\begin{align}
\biggl|\frac{t+1}{1-t}\biggr|=\biggl|\frac{t'+1}{1-t'}\biggr|&\iff \bigl|(1+t)(1-t')\bigr|=\bigl|(1+t')(1-t)\bigr| \\
&\iff \bigl|1+t-t'-tt'\bigr|=\bigl|1+t'-t-tt'\bigr|\\
&\iff \bigl|t-t'\bigr|=\bigl|t'-t\bigr|.
\end{align}
A: The first equation is a degenerate conic that represents a pair of distinct lines when $h^2\gt a^2$. The associated matrix is $\small{\begin{bmatrix}a&h\\h&a\end{bmatrix}}$, and we can find by inspection the eigenvectors $(1,1)^T$ and $(-1,1)^T$. Thus, the conic is symmetric about the line $x=y$, which is perpendicular to $x+y=5$. The triangle formed by the three pairwise intersection points of the lines is therefore itself symmetric about $x=y$, i.e., it is isosceles.
