If the line (2x + 3y + 4) + lambda(6x - y + 1 | vedclass

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(B) The given line is $(2x + 3y + 4) + \lambda(6x - y + 12) = 0$. Rearranging the terms,we get $(2 + 6\lambda)x + (3 - \lambda)y + (4 + 12\lambda) = 0

Vedclass · Accepted Answer

$\frac{-29}{37}$

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(B) The given line is $(2x + 3y + 4) + \lambda(6x - y + 12) = 0$. Rearranging the terms,we get $(2 + 6\lambda)x + (3 - \lambda)y + (4 + 12\lambda) = 0$. The slope of this line $(m_1)$ is $-\frac{2 + 6\lambda}{3 - \lambda}$. The second line is $7x + 5y = 2$,which can be written as $5y = -7x + 2$,or $y = -\frac{7}{5}x + \frac{2}{5}$. The slope of this line $(m_2)$ is $-\frac{7}{5}$. Since the lines are perpendicular,the product of their slopes is $-1$,so $m_1 	imes m_2 = -1$. $(-\frac{2 + 6\lambda}{3 - \lambda}) 	imes (-\frac{7}{5}) = -1$. $\frac{14 + 42\lambda}{5(3 - \lambda)} = -1$. $14 + 42\lambda = -5(3 - \lambda)$. $14 + 42\lambda = -15 + 5\lambda$. $42\lambda - 5\lambda = -15 - 14$. $37\lambda = -29$. $\lambda = -\frac{29}{37}$.

If the line $(2x + 3y + 4) + \lambda(6x - y + 12) = 0$ is perpendicular to the line $7x + 5y = 2$,then $\lambda = $

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