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QuestionFind the equation of the line passsing through the point P $\left(8,\frac{1}{5}\right)$ and perpendicular to the line $2x-3y-1= 0$
AnswerThe slope of the line $2x-3y-1= 0$ is given by $m = \frac{2}{3}$
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 $m=-\frac{3}{2}$
The second line passes through the point $\left(8,\frac{1}{5}\right)$ and has slope $m=-\frac{3}{2}$
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 $m=-\frac{3}{2}$, $x_0=8$ and $y_0=\frac{1}{5}$
Substituting these values in the equation, we get
$ y-\frac{1}{5}=-\frac{3}{2}\left(x-8\right)$
or, $y-\frac{1}{5}=-\frac{3}{2}x+12$
or, $y=-\frac{3}{2}x+\frac{61}{5}$