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If the lines $\mathrm{a x^2+2 h x y+b y^2=0}$ be two sides of a parallelogram and the line $\mathrm{lx + my = 1}$ be one of its diagonals, then the equation of the other diagonal is $\mathrm{y(bl - hm) = kx}$ , where k =
Option: 1
$\mathrm{am-hl}$

Option: 2
$\mathrm{am+hl}$

Option: 3
$\mathrm{\left ( ah+ml \right )}$

Option: 4
$\mathrm{\left ( ah-ml \right )}$

Answers (1)

best_answer

Let OABC be a parallelogram then equation of OA and OB is

$\mathrm{a x^2+b y^2+2 h x y=0}\: \: \: \: \: \: \: \: \: \: \: ...(1)$

Equation of AB is $\mathrm{lx+my=1}$ ........(2)

From equation (1) and (2)

$\mathrm{a x^2+b\left\{\frac{1-I x}{m}\right\}^2+2 h x\left\{\frac{1-I x}{m}\right\}=0}$

$\mathrm{\Rightarrow x^2\left(a m^2+\left.b\right|^2-2 h l m\right)+x(2 h m-2 l b)+b=0}$

Let $\mathrm{\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { and }\left(\mathrm{x}_2, \mathrm{y}_2\right)}$ be the coordinates of point A and B

Then from equation of (3)

Sum of the roots $\mathrm{x_1+x_2=\frac{2 l b-2 h m}{a m^2+bl^2-2 h l m}}$

So abscissa of the point D i.e. $\mathrm{\frac{\mathrm{x}_1+\mathrm{x}_2}{2}=\frac{\mathrm{lb}-\mathrm{hm}}{\mathrm{am^{2 } + \mathrm { b }} \mathrm{l}^2-2 \mathrm{hlm}}}$

Ordinate of the mid-point D i.e. $\mathrm{\frac{y_1+y_2}{2}=\frac{1-l\left(\frac{x_1+x_2}{2}\right)}{m}}$

$\mathrm{=\frac{a m-l h}{a m^2+bl^2-2 h l m}}$

so equation of the other diagonal passing through the points O and D will be P

$\begin{aligned} & \mathrm{(y-0)=\frac{a m-l h}{l b-h m}(x-0) }\\ \\& \Rightarrow \mathrm{y(b l-h m)=x(a m-h l)} \end{aligned}$

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