Equation of a line (point/parallel line)

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Find the equation of the line passsing through the point P $\left(\frac{-1}{3},\frac{-3}{7}\right)$ and perpendicular to the line $3x-5y+6= 0$

**Answer**

The slope of the line $3x-5y+6= 0$ is given by $m = \frac{3}{5}$

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 $m=-\frac{5}{3}$

The second line passes through the point $\left(\frac{-1}{3},\frac{-3}{7}\right)$ and has slope $m=-\frac{5}{3}$

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 $m=-\frac{5}{3}$, $x_0=-\frac{1}{3}$ and $y_0=-\frac{3}{7}$

Substituting these values in the equation, we get

$ y-\left(-\frac{3}{7}\right)=-\frac{5}{3}\left(x-\left(-\frac{1}{3}\right)\right)$

or, $ y+\frac{3}{7}=-\frac{5}{3}\left(x+\frac{1}{3}\right)$

or, $y+\frac{3}{7}=-\frac{5}{3}x-\frac{5}{9}$

or, $y=-\frac{5}{3}x-\frac{62}{63}$