Equation of a line (point/perpendicular line)

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Find the equation of the line passsing through the point P $\left(\frac{-7}{9},-1\right)$ and parallel to the line $x+7y-10= 0$

**Answer**

The slope of the line $x+7y-10= 0$ is given by $m = -\frac{1}{7}$

The second line is parallel to this line. So, they have the same slopes $m=-\frac{1}{7}$

The second line passes through the point $\left(\frac{-7}{9},-1\right)$ and has slope $m=-\frac{1}{7}$

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 $m=-\frac{1}{7}$, $x_0=-\frac{7}{9}$ and $y_0=-1$

Substituting these values in the equation, we get

$ y-\left(-1\right)=-\frac{1}{7}\left(x-\left(-\frac{7}{9}\right)\right)$

or, $ y+1=-\frac{1}{7}\left(x+\frac{7}{9}\right)$

or, $y+1=-\frac{1}{7}x-\frac{1}{9}$

or, $y=-\frac{1}{7}x-\frac{10}{9}$