<s>[INST] coordinate geometry high level problems 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 $Y(bl – lm) = X(am – hl)$
 [/INST] 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
$$\frac{Y}{X}=\frac{y_1+y_2}{x_1+x_2}$$
$$(b\ell-hm)Y=(am-h\ell)X$$
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