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

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

Vedclass · Accepted Answer

$3x + 4y = 1$

Vedclass · Answer

(D) Let $P = \left(-\frac{7}{5}, -\frac{6}{5}\right)$ and $Q = (1, 2)$. The line is the perpendicular bisector of the segment $PQ$.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 $m_{PQ} = \frac{2 - (-\frac{6}{5})}{1 - (-\frac{7}{5})} = \frac{\frac{16}{5}}{\frac{12}{5}} = \frac{4}{3}$.The slope of the line perpendicular to $PQ$ is $m = -\frac{1}{m_{PQ}} = -\frac{3}{4}$.The equation of the line passing through $M$ with slope $m$ is $y - \frac{2}{5} = -\frac{3}{4}\left(x + \frac{1}{5}\right)$.Multiplying by $20$: $20y - 8 = -15x - 3$,which simplifies to $15x + 20y = 5$,or $3x + 4y = 1$.

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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