Equation of a line (point/perpendicular line)

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Find the equation of the line passsing through the point P $\left(\frac{10}{3},\frac{-1}{2}\right)$ and parallel to the line $x+4y+4= 0$

**Answer**

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

The second line is parallel to this line. So, they have the same slopes $m=-\frac{1}{4}$

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

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

Substituting these values in the equation, we get

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

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

or, $y+\frac{1}{2}=-\frac{1}{4}x+\frac{5}{6}$

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