Equation of a line (point/point of intersection)

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Find the equation of the line passsing through the point P $\left(\frac{-9}{7},-9\right)$ and the intersection of the lines $7x+8y-1= 0$ and $3y-1= 0$

**Answer**

The lines $7x+8y-1= 0$ and $3y-1= 0$ intersect at the point Q$\left(\frac{5}{21},\frac{-1}{3}\right)$

The given line passes through the point P $\left(\frac{-9}{7},-9\right)$ and $\left(\frac{5}{21},\frac{-1}{3}\right)$

The slope of the line joining the points $P\left(\frac{-9}{7},-9\right)$ and $Q\left(\frac{5}{21},\frac{-1}{3}\right)$ is given by $m=\frac{\frac{-1}{3}+9}{\frac{5}{21}+\frac{9}{7}}=\frac{91}{16}$

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

$y =\frac{91}{16}x-9-\left(\frac{91}{16}\right)\left(-\frac{9}{7}\right)$

or, $y =\frac{91}{16}x-9+\frac{117}{16}$

or, $y=\frac{91}{16}x-\frac{27}{16}$