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

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(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $

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$2$

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(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$. The point of intersection $(x_0, y_0)$ of the pair of lines is given by the formula: $x_0 = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-36} = \frac{-48}{-36} = \frac{4}{3}$. $y_0 = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$. Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the line $5x+\lambda y-8=0$. $5(\frac{4}{3}) + \lambda(\frac{2}{3}) - 8 = 0$. $\frac{20}{3} + \frac{2\lambda}{3} - 8 = 0$. Multiply by $3$: $20 + 2\lambda - 24 = 0$. $2\lambda - 4 = 0$. $2\lambda = 4$. $\lambda = 2$.

If the lines $x^2+2xy-35y^2-4x+44y-12=0$ and $5x+\lambda y-8=0$ are concurrent,then the value of $\lambda$ is

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