Equation of a line (point/parallel line)

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

**Answer**

The slope of the line $x+y+7= 0$ is given by $m = -1$

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 $m=1$

The second line passes through the point $\left(-8,\frac{-1}{6}\right)$ and has slope $m=1$

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 $m=1$, $x_0=-8$ and $y_0=-\frac{1}{6}$

Substituting these values in the equation, we get

$ y-\left(-\frac{1}{6}\right)=\left(x-\left(-8\right)\right)$

or, $ y+\frac{1}{6}=\left(x+8\right)$

or, $y+\frac{1}{6}=x+8$

or, $y=x+\frac{47}{6}$