Equation of a line (point/parallel line)

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Find the equation of the line passsing through the point P $\left(-1,3\right)$ and perpendicular to the line $5x-8y-9= 0$

**Answer**

The slope of the line $5x-8y-9= 0$ is given by $m = \frac{5}{8}$

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 $m=-\frac{8}{5}$

The second line passes through the point $\left(-1,3\right)$ and has slope $m=-\frac{8}{5}$

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 $m=-\frac{8}{5}$, $x_0=-1$ and $y_0=3$

Substituting these values in the equation, we get

$ y-\left(3\right)=-\frac{8}{5}\left(x-\left(-1\right)\right)$

or, $ y-3=-\frac{8}{5}\left(x+1\right)$

or, $y-3=-\frac{8}{5}x-\frac{8}{5}$

or, $y=-\frac{8}{5}x+\frac{7}{5}$