Equation of a line (point/perpendicular line)

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Find the equation of the line passsing through the point P $\left(\frac{1}{5},\frac{5}{2}\right)$ and parallel to the line $x-3y-4= 0$

**Answer**

The slope of the line $x-3y-4= 0$ is given by $m = \frac{1}{3}$

The second line is parallel to this line. So, they have the same slopes $m=\frac{1}{3}$

The second line passes through the point $\left(\frac{1}{5},\frac{5}{2}\right)$ and has slope $m=\frac{1}{3}$

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 $m=\frac{1}{3}$, $x_0=\frac{1}{5}$ and $y_0=\frac{5}{2}$

Substituting these values in the equation, we get

$ y-\frac{5}{2}=\frac{1}{3}\left(x-\frac{1}{5}\right)$

or, $y-\frac{5}{2}=\frac{1}{3}x-\frac{1}{15}$

or, $y=\frac{1}{3}x+\frac{73}{30}$