Equation of a line (point/parallel line)

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Find the equation of the line passsing through the point P $\left(\frac{-1}{2},\frac{1}{10}\right)$ and perpendicular to the line $8x+6y+1= 0$

**Answer**

The slope of the line $8x+6y+1= 0$ is given by $m = -\frac{4}{3}$

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 $m=\frac{3}{4}$

The second line passes through the point $\left(\frac{-1}{2},\frac{1}{10}\right)$ and has slope $m=\frac{3}{4}$

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 $m=\frac{3}{4}$, $x_0=-\frac{1}{2}$ and $y_0=\frac{1}{10}$

Substituting these values in the equation, we get

$ y-\left(\frac{1}{10}\right)=\frac{3}{4}\left(x-\left(-\frac{1}{2}\right)\right)$

or, $ y-\frac{1}{10}=\frac{3}{4}\left(x+\frac{1}{2}\right)$

or, $y-\frac{1}{10}=\frac{3}{4}x+\frac{3}{8}$

or, $y=\frac{3}{4}x+\frac{19}{40}$