Equation of a line (point/perpendicular line)

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Find the equation of the line passsing through the point P $\left(\frac{-5}{4},\frac{1}{5}\right)$ and parallel to the line $7x+6y-8= 0$

**Answer**

The slope of the line $7x+6y-8= 0$ is given by $m = -\frac{7}{6}$

The second line is parallel to this line. So, they have the same slopes $m=-\frac{7}{6}$

The second line passes through the point $\left(\frac{-5}{4},\frac{1}{5}\right)$ and has slope $m=-\frac{7}{6}$

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 $m=-\frac{7}{6}$, $x_0=-\frac{5}{4}$ and $y_0=\frac{1}{5}$

Substituting these values in the equation, we get

$ y-\left(\frac{1}{5}\right)=-\frac{7}{6}\left(x-\left(-\frac{5}{4}\right)\right)$

or, $ y-\frac{1}{5}=-\frac{7}{6}\left(x+\frac{5}{4}\right)$

or, $y-\frac{1}{5}=-\frac{7}{6}x-\frac{35}{24}$

or, $y=-\frac{7}{6}x-\frac{151}{120}$