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If the lines $x^2+2 x y-35 y^2-4 x+44 y-12=0$ and $5 x+\lambda y-8=0$ are concurrent, then the value of $\lambda$ is
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The correct answer is:
$2$
Given line is
$x^2+2 x y-35 y^2-4 x+44 y-12=0$
Here, $a=1, b=-35, c=-12, h=1, f=22$, $g=-2$
$\therefore$ Point of intersection is $\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}\right)$
$=\left(\frac{22-70}{-35-1}, \frac{-2-22}{-35-1}\right)$
$\begin{aligned} & =\left(\frac{48}{36}, \frac{24}{36}\right) \\ & =\left(\frac{4}{3}, \frac{2}{3}\right)\end{aligned}$
If the lines are concurrent
Then, $5\left(\frac{4}{3}\right)+\lambda\left(\frac{2}{3}\right)-8=0$
$\begin{array}{ll}\Rightarrow & \frac{2}{3} \lambda=8-\frac{20}{3} \\ \Rightarrow & \lambda=\frac{24-20}{3} \times \frac{3}{2} \\ \Rightarrow & \lambda=2\end{array}$
$x^2+2 x y-35 y^2-4 x+44 y-12=0$
Here, $a=1, b=-35, c=-12, h=1, f=22$, $g=-2$
$\therefore$ Point of intersection is $\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}\right)$
$=\left(\frac{22-70}{-35-1}, \frac{-2-22}{-35-1}\right)$
$\begin{aligned} & =\left(\frac{48}{36}, \frac{24}{36}\right) \\ & =\left(\frac{4}{3}, \frac{2}{3}\right)\end{aligned}$
If the lines are concurrent
Then, $5\left(\frac{4}{3}\right)+\lambda\left(\frac{2}{3}\right)-8=0$
$\begin{array}{ll}\Rightarrow & \frac{2}{3} \lambda=8-\frac{20}{3} \\ \Rightarrow & \lambda=\frac{24-20}{3} \times \frac{3}{2} \\ \Rightarrow & \lambda=2\end{array}$
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