Equation of a line (point/point of intersection)

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Find the equation of the line passsing through the point P $\left(\frac{-2}{3},1\right)$ and the intersection of the lines $4x-10y+3= 0$ and $9x+8y= 0$

**Answer**

The lines $4x-10y+3= 0$ and $9x+8y= 0$ intersect at the point Q$\left(\frac{12}{61},\frac{-27}{122}\right)$

The given line passes through the point P $\left(\frac{-2}{3},1\right)$ and $\left(\frac{12}{61},\frac{-27}{122}\right)$

The slope of the line joining the points $P\left(\frac{-2}{3},1\right)$ and $Q\left(\frac{12}{61},\frac{-27}{122}\right)$ is given by $m=\frac{\frac{-27}{122}-1}{\frac{12}{61}+\frac{2}{3}}=\frac{-447}{316}$

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

$y =-\frac{447}{316}x+-\left(-\frac{447}{316}\right)\left(-\frac{2}{3}\right)$

or, $y =-\frac{447}{316}x+1-\frac{149}{158}$

or, $y=-\frac{447}{316}x+\frac{9}{158}$