If the lines x^2+2xy-35y^2-4x+44y-12=0 and 5x | vedclass

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(C) The given pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-1

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$2$

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(C) The given pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.The point of intersection $(x, y)$ of the pair of lines is given by the formula $\left(\frac{hf-bg}{ab-h^2}, \frac{gh-af}{ab-h^2}\right)$.Substituting the values: $x = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-36} = \frac{-48}{-36} = \frac{4}{3}$.$y = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$.The point of intersection is $(\frac{4}{3}, \frac{2}{3})$.Since the lines are concurrent,this point must satisfy the equation $5x+ky-8=0$.$5(\frac{4}{3}) + k(\frac{2}{3}) - 8 = 0$.$\frac{20}{3} + \frac{2k}{3} = 8$.$20 + 2k = 24$.$2k = 4 \Rightarrow k = 2$.

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

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