coordinate geometry high level problems

Question

If the lines $aX^2 + 2hXY + bY^2 =0$ form two sides of a parallelogram and the line $lX + mY =1$ is one diagonal, prove that the equation of other diagonal is $Y(bl – lm) = X(am – hl)$

Answer 1

Let the pair of straight lines meet the diagonal at $(x_1,y_1)$ and $(x_2,y_2)$, then the other diagonal will pass through $(0,0)$ and $(x_1+x_2,y_1+y_2)$.

$$ \left \{ \begin{align*} aX^2+2hXY+bY^2 &= 0 \\ \ell X+mY &= 1 \end{align*} \right.$$

$$\implies aX^2+2hX\left( \frac{1-\ell X}{m} \right)+ b\left( \frac{1-\ell X}{m} \right)^2=0$$

$$\left( a-\frac{2h\ell}{m}+\frac{b\ell^2}{m^2} \right)X^2+ \left( \frac{2h}{m}-\frac{2b\ell}{m^2} \right)X+\frac{b}{m^2}=0$$

$$x_1+x_2=\frac{2(b\ell-hm)}{am^2-2h\ell m+bl^2}$$

Similarly, $$y_1+y_2=\frac{2(am-h\ell)}{am^2-2h\ell m+bl^2}$$

Hence the equation for the other diagonal is

$$\frac{Y}{X}=\frac{y_1+y_2}{x_1+x_2}$$

$$(b\ell-hm)Y=(am-h\ell)X$$