If the image of left(frac{-7}{5}, frac{-6}{5} | vedclass

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(D) Let the point $P = \left(-\frac{7}{5}, -\frac{6}{5}\right)$ and its image $Q = (1, 2)$. The line is the perpendicular bisector of $PQ$.The midpoin

Vedclass · Accepted Answer

$3x + 4y = 1$

Vedclass · Answer

(D) Let the point $P = \left(-\frac{7}{5}, -\frac{6}{5}\right)$ and its image $Q = (1, 2)$. The line is the perpendicular bisector of $PQ$.The midpoint $M$ of $PQ$ is $\left(\frac{-\frac{7}{5} + 1}{2}, \frac{-\frac{6}{5} + 2}{2}\right) = \left(-\frac{1}{5}, \frac{2}{5}\right)$.The slope of $PQ$ is $m_{PQ} = \frac{2 - (-6/5)}{1 - (-7/5)} = \frac{16/5}{12/5} = \frac{4}{3}$.The slope of the required line is $m = -\frac{1}{m_{PQ}} = -\frac{3}{4}$.Using the point-slope form $y - y_1 = m(x - x_1)$ at point $M$:$y - \frac{2}{5} = -\frac{3}{4}\left(x + \frac{1}{5}\right)$$\frac{5y - 2}{5} = -\frac{3}{4} \cdot \frac{5x + 1}{5}$$4(5y - 2) = -3(5x + 1)$$20y - 8 = -15x - 3$$15x + 20y = 5$Dividing by $5$,we get $3x + 4y = 1$.

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

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