If the two lines frac{3}{2} x + (2a - 1)y = 3 | vedclass

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(B) Given that the two lines $L_1: \frac{3}{2}x + (2a - 1)y = 3$ and $L_2: 2x + a^2y = -3$ are perpendicular.Since $L_1 \perp L_2$,the product of thei

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$3$

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(B) Given that the two lines $L_1: \frac{3}{2}x + (2a - 1)y = 3$ and $L_2: 2x + a^2y = -3$ are perpendicular.Since $L_1 \perp L_2$,the product of their slopes $m_1 	imes m_2 = -1$.The slope of $L_1$ is $m_1 = -\frac{3/2}{2a - 1} = -\frac{3}{4a - 2}$.The slope of $L_2$ is $m_2 = -\frac{2}{a^2}$.Thus,$(-\frac{3}{4a - 2}) 	imes (-\frac{2}{a^2}) = -1 \Rightarrow \frac{6}{a^2(4a - 2)} = -1$.$6 = -4a^3 + 2a^2$ $\Rightarrow 4a^3 - 2a^2 + 6 = 0$ $\Rightarrow 2a^3 - a^2 + 3 = 0$.Testing $a = -1$: $2(-1)^3 - (-1)^2 + 3 = -2 - 1 + 3 = 0$. So,$(a + 1)$ is a factor.Dividing $2a^3 - a^2 + 3$ by $(a + 1)$ gives $2a^2 - 3a + 3 = 0$,which has no real roots $(D = 9 - 24 So,$a = -1$.Substituting $a = -1$ into the lines:$L_1: \frac{3}{2}x + (2(-1) - 1)y = 3$ $\Rightarrow \frac{3}{2}x - 3y = 3$ $\Rightarrow x - 2y = 2$.$L_2: 2x + (-1)^2y = -3 \Rightarrow 2x + y = -3$.Solving the system:$x - 2y = 2$ $(i)$$2x + y = -3$ (ii)From (ii),$y = -3 - 2x$. Substituting into $(i)$: $x - 2(-3 - 2x) = 2$ $\Rightarrow x + 6 + 4x = 2$ $\Rightarrow 5x = -4$ $\Rightarrow x = -\frac{4}{5}$.Then $y = -3 - 2(-\frac{4}{5}) = -3 + \frac{8}{5} = -\frac{7}{5}$.The intersection point is $(-\frac{4}{5}, -\frac{7}{5})$.The distance from $(1, 1)$ is $\sqrt{(1 - (-\frac{4}{5}))^2 + (1 - (-\frac{7}{5}))^2} = \sqrt{(\frac{9}{5})^2 + (\frac{12}{5})^2} = \sqrt{\frac{81 + 144}{25}} = \sqrt{\frac{225}{25}} = \sqrt{9} = 3$.

If the two lines $\frac{3}{2} x + (2a - 1)y = 3$ and $2x + a^2y = -3$ are perpendicular,then the distance of their point of intersection from the point $(1, 1)$ is

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