If straight lines alpha^2x + alpha y = 9 and | vedclass

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Question

(A) The equation of the first line is $\alpha^2x + \alpha y = 9$,which can be written as $\alpha y = -\alpha^2x + 9$,or $y = -\alpha x + \frac{9}{\alp

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$-2/3$

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(A) The equation of the first line is $\alpha^2x + \alpha y = 9$,which can be written as $\alpha y = -\alpha^2x + 9$,or $y = -\alpha x + \frac{9}{\alpha}$. The slope $m_1$ is $-\alpha$.The equation of the second line is $3x + 2y = 5$,which can be written as $2y = -3x + 5$,or $y = -\frac{3}{2}x + \frac{5}{2}$. The slope $m_2$ is $-\frac{3}{2}$.Since the lines are perpendicular,the product of their slopes must be $-1$,so $m_1 	imes m_2 = -1$.Substituting the values,$(-\alpha) 	imes (-\frac{3}{2}) = -1$.$\frac{3}{2}\alpha = -1$.$\alpha = -\frac{2}{3}$.

If straight lines $\alpha^2x + \alpha y = 9$ and $3x + 2y = 5$ are perpendicular,then the value of $\alpha$ is

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