Equation of a line (point/parallel line)
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Question

Find the equation of the line passsing through the point P $\left(6,\frac{-2}{7}\right)$ and perpendicular to the line $7x+10y+7= 0$

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

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 $m=\frac{10}{7}$

The second line passes through the point $\left(6,\frac{-2}{7}\right)$ and has slope $m=\frac{10}{7}$

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 $m=\frac{10}{7}$, $x_0=6$ and $y_0=-\frac{2}{7}$

Substituting these values in the equation, we get

$y-\left(-\frac{2}{7}\right)=\frac{10}{7}\left(x-\left(6\right)\right)$

or, $y+\frac{2}{7}=\frac{10}{7}\left(x-6\right)$

or, $y+\frac{2}{7}=\frac{10}{7}x-\frac{60}{7}$

or, $y=\frac{10}{7}x-\frac{62}{7}$