The locus of the image of a variable point P( | vedclass

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(A) Let the image of point $P(\alpha, 2 \alpha-1)$ with respect to the line $3 x-2 y+4=0$ be $(x, y)$.Using the formula for the image of a point $(x_1

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$22(13 x+36)=19(13 y-11)$

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(A) Let the image of point $P(\alpha, 2 \alpha-1)$ with respect to the line $3 x-2 y+4=0$ be $(x, y)$.Using the formula for the image of a point $(x_1, y_1)$ with respect to the line $ax+by+c=0$ is $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2(ax_1+by_1+c)}{a^2+b^2}$.Here,$x_1=\alpha, y_1=2 \alpha-1, a=3, b=-2, c=4$.Substituting these values:$\frac{x-\alpha}{3}=\frac{y-(2 \alpha-1)}{-2}=\frac{-2(3(\alpha)-2(2 \alpha-1)+4)}{3^2+(-2)^2}$$\frac{x-\alpha}{3}=\frac{y-2 \alpha+1}{-2}=\frac{-2(3 \alpha-4 \alpha+2+4)}{9+4}$$\frac{x-\alpha}{3}=\frac{y-2 \alpha+1}{-2}=\frac{-2(-\alpha+6)}{13} = \frac{2 \alpha-12}{13}$Now,equate the parts to find $x$ and $y$ in terms of $\alpha$:$1) \frac{x-\alpha}{3} = \frac{2 \alpha-12}{13} \implies 13x - 13\alpha = 6\alpha - 36 \implies 13x + 36 = 19\alpha \implies \alpha = \frac{13x+36}{19}$$2) \frac{y-2\alpha+1}{-2} = \frac{2 \alpha-12}{13} \implies 13y - 26\alpha + 13 = -4\alpha + 24 \implies 13y - 11 = 22\alpha \implies \alpha = \frac{13y-11}{22}$Equating the two expressions for $\alpha$:$\frac{13x+36}{19} = \frac{13y-11}{22}$$22(13x+36) = 19(13y-11)$

The locus of the image of a variable point $P(\alpha, 2 \alpha-1)$ with respect to the line $3 x-2 y+4=0$ is

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