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QuestionFind the equation of the line passsing through the point P $\left(\frac{2}{7},\frac{-5}{4}\right)$ and perpendicular to the line $3x-8y-6= 0$
AnswerThe slope of the line $3x-8y-6= 0$ is given by $m = \frac{3}{8}$
The second line is perpendicular to this line.
So, the product of their slopes must be $-1$. Thus, the slope of the second line must me $m=-\frac{8}{3}$
The second line passes through the point $\left(\frac{2}{7},\frac{-5}{4}\right)$ and has slope $m=-\frac{8}{3}$
The equation of the line with slope $m$ and passing through the point $\left(x_0,y_0\right)$ is given by $$y-y_0=m\left(x-x_0\right)$$
Here $m=-\frac{8}{3}$, $x_0=\frac{2}{7}$ and $y_0=-\frac{5}{4}$
Substituting these values in the equation, we get
$ y-\left(-\frac{5}{4}\right)=-\frac{8}{3}\left(x-\left(\frac{2}{7}\right)\right)$
or, $ y+\frac{5}{4}=-\frac{8}{3}\left(x-\frac{2}{7}\right)$
or, $y+\frac{5}{4}=-\frac{8}{3}x+\frac{16}{21}$
or, $y=-\frac{8}{3}x-\frac{41}{84}$