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If the image of $\left(\frac{-7}{5}, \frac{-6}{5}\right)$ in a line is $(1, 2)$,then the equation of the line is

Assuming that straight lines work as the plane mirror for a point,find the image of the point $(1, 2)$ in the line $x - 3y + 4 = 0$.

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If the image of point $P(2, 3)$ in a line $L$ is $Q(4, 5)$,then the image of point $R(0, 0)$ in the same line is

The image of the point $(3,5)$ in the line $x-y+1=0$ lies on:

The length of the perpendicular from the origin,on the normal to the curve $x^{2}+2xy-3y^{2}=0$ at the point $(2,2)$ is

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