Equation of a line (point/point of intersection)

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Find the equation of the line passsing through the point P $\left(1,\frac{-7}{6}\right)$ and the intersection of the lines $4x+4y+5= 0$ and $10x+5y+7= 0$

**Answer**

The lines $4x+4y+5= 0$ and $10x+5y+7= 0$ intersect at the point Q$\left(\frac{3}{20},\frac{11}{10}\right)$

The given line passes through the point P $\left(1,\frac{-7}{6}\right)$ and $\left(\frac{3}{20},\frac{11}{10}\right)$

The slope of the line joining the points $P\left(1,\frac{-7}{6}\right)$ and $Q\left(\frac{3}{20},\frac{11}{10}\right)$ is given by $m=\frac{\frac{11}{10}+\frac{7}{6}}{\frac{3}{20}-1}=\frac{-8}{3}$

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

$y =-\frac{8}{3}x-\frac{7}{6}-\left(-\frac{8}{3}\right)\left(1\right)$

or, $y =-\frac{8}{3}x-\frac{7}{6}+\frac{8}{3}$

or, $y=-\frac{8}{3}x+\frac{3}{2}$