Equation of a line (point/point of intersection)

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Find the equation of the line passsing through the point P $\left(\frac{1}{6},\frac{-2}{7}\right)$ and the intersection of the lines $3x+8y-8= 0$ and $7x-6y+2= 0$

**Answer**

The lines $3x+8y-8= 0$ and $7x-6y+2= 0$ intersect at the point Q$\left(\frac{-16}{37},\frac{-31}{37}\right)$

The given line passes through the point P $\left(\frac{1}{6},\frac{-2}{7}\right)$ and $\left(\frac{-16}{37},\frac{-31}{37}\right)$

The slope of the line joining the points $P\left(\frac{1}{6},\frac{-2}{7}\right)$ and $Q\left(\frac{-16}{37},\frac{-31}{37}\right)$ is given by $m=\frac{\frac{-31}{37}+\frac{2}{7}}{\frac{-16}{37}-\frac{1}{6}}=\frac{858}{931}$

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

$y =\frac{858}{931}x-\frac{2}{7}-\left(\frac{858}{931}\right)\left(\frac{1}{6}\right)$

or, $y =\frac{858}{931}x-\frac{2}{7}-\frac{143}{931}$

or, $y=\frac{858}{931}x-\frac{409}{931}$