Equation of a line (point/parallel line)

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Find the equation of the line passsing through the point P $\left(\frac{7}{9},\frac{-9}{8}\right)$ and perpendicular to the line $7x+5y-4= 0$

**Answer**

The slope of the line $7x+5y-4= 0$ is given by $m = -\frac{7}{5}$

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 $m=\frac{5}{7}$

The second line passes through the point $\left(\frac{7}{9},\frac{-9}{8}\right)$ and has slope $m=\frac{5}{7}$

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 $m=\frac{5}{7}$, $x_0=\frac{7}{9}$ and $y_0=-\frac{9}{8}$

Substituting these values in the equation, we get

$ y-\left(-\frac{9}{8}\right)=\frac{5}{7}\left(x-\left(\frac{7}{9}\right)\right)$

or, $ y+\frac{9}{8}=\frac{5}{7}\left(x-\frac{7}{9}\right)$

or, $y+\frac{9}{8}=\frac{5}{7}x-\frac{5}{9}$

or, $y=\frac{5}{7}x-\frac{121}{72}$