If a line ax + 2y = k forms a triangle of are | vedclass

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(A) The line $ax + 2y = k$ is perpendicular to $2x - 3y + 7 = 0$. The slope of $ax + 2y = k$ is $m_1 = -a/2$. The slope of $2x - 3y + 7 = 0$ is $m_2 =

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$-36$

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(A) The line $ax + 2y = k$ is perpendicular to $2x - 3y + 7 = 0$. The slope of $ax + 2y = k$ is $m_1 = -a/2$. The slope of $2x - 3y + 7 = 0$ is $m_2 = 2/3$. Since they are perpendicular,$m_1 	imes m_2 = -1$. $(-a/2) 	imes (2/3) = -1 \implies -a/3 = -1 \implies a = 3$. The equation of the line is $3x + 2y = k$. Converting to intercept form: $\frac{x}{k/3} + \frac{y}{k/2} = 1$. The area of the triangle formed with coordinate axes is $\frac{1}{2} 	imes |\frac{k}{3}| 	imes |\frac{k}{2}| = 3$. $\frac{|k^2|}{12} = 3 \implies |k^2| = 36 \implies k^2 = 36 \implies k = \pm 6$. The product of all possible values of $k$ is $6 	imes (-6) = -36$.

If a line $ax + 2y = k$ forms a triangle of area $3$ sq. units with the coordinate axes and is perpendicular to the line $2x - 3y + 7 = 0$,then the product of all the possible values of $k$ is

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