Equation of a line (point/point of intersection)

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

**Answer**

The lines $6x+4y+1= 0$ and $x-7y+4= 0$ intersect at the point Q$\left(\frac{1}{2},\frac{-1}{2}\right)$

The given line passes through the point P $\left(-1,\frac{-1}{7}\right)$ and $\left(\frac{1}{2},\frac{-1}{2}\right)$

The slope of the line joining the points $P\left(-1,\frac{-1}{7}\right)$ and $Q\left(\frac{1}{2},\frac{-1}{2}\right)$ is given by $m=\frac{\frac{-1}{2}+\frac{1}{7}}{\frac{1}{2}+1}=\frac{-5}{21}$

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

$y =-\frac{5}{21}x-\frac{1}{7}-\left(-\frac{5}{21}\right)\left(-1\right)$

or, $y =-\frac{5}{21}x-\frac{1}{7}-\frac{5}{21}$

or, $y=-\frac{5}{21}x-\frac{8}{21}$