If the lines $x^2+2 x y-35 y^2-4 x+44 y-12=0$ and $5 x+k y-8=0$ are concurrent, then $k$ equals

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Aicharya · Accepted Answer

Since, we know that, point of intersection of pairs of line represented by $\begin{aligned} & a x^2+2 h y x+b y^2+2 g x+2 f y+c=0 	ext { is } \ & \qquad\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}ight)\end{aligned}$ For the lines, $x^2+2 x y-35 y^2-4 x+44 y-12=0$ $a=1, h=1, b=-35, g=-2, f=22$ and $c=-12$ $\begin{aligned} & 	herefore\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}ight) \ & =\left(\frac{1 	imes 22-70}{-35-1}, \frac{-2-22}{-35-1}ight)=\left(\frac{-48}{-36}, \frac{-24}{-36}ight)=\left(\frac{4}{3}, \frac{2}{3}ight)\end{aligned}$ $\because$ Lines are concurrent, hence $ \begin{array}{rlrl} \left(\frac{4}{3}, \frac{2}{3}ight) 	ext { must satisfy } 5 x+k y-8 & =0 \ 5\left(\frac{4}{3}ight)+k\left(\frac{2}{3}ight) & =8 \ 20+2 k & =24 \Rightarrow 2 k=4 \ \Rightarrow & k & =2 \end{array} $

If the lines $x^2+2 x y-35 y^2-4 x+44 y-12=0$ and $5 x+k y-8=0$ are concurrent, then $k$ equals

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