Equation of a line (point/point of intersection)

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Find the equation of the line passsing through the point P $\left(-3,\frac{-1}{2}\right)$ and the intersection of the lines $4x-4y+1= 0$ and $7x+2y+2= 0$

**Answer**

The lines $4x-4y+1= 0$ and $7x+2y+2= 0$ intersect at the point Q$\left(\frac{5}{18},\frac{1}{36}\right)$

The given line passes through the point P $\left(-3,\frac{-1}{2}\right)$ and $\left(\frac{5}{18},\frac{1}{36}\right)$

The slope of the line joining the points $P\left(-3,\frac{-1}{2}\right)$ and $Q\left(\frac{5}{18},\frac{1}{36}\right)$ is given by $m=\frac{\frac{1}{36}+\frac{1}{2}}{\frac{5}{18}+3}=\frac{19}{118}$

The equation of the line is$$y - y_0 = m\left(x - x_0\right)$$

or, $$y= mx + y_0-mx_0$$

Substituting the values of $m$, $x_0$ and $y_0$, we get

$y =\frac{19}{118}x-\frac{1}{2}-\left(\frac{19}{118}\right)\left(-3\right)$

or, $y =\frac{19}{118}x-\frac{1}{2}+\frac{57}{118}$

or, $y=\frac{19}{118}x-\frac{1}{59}$