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QuestionFind the equation of the line passsing through the point P $\left(1,3\right)$ and the intersection of the lines $9x+4y-6= 0$ and $2x+2y+1= 0$
AnswerThe lines $9x+4y-6= 0$ and $2x+2y+1= 0$ intersect at the point Q$\left(\frac{-8}{5},\frac{21}{10}\right)$
The given line passes through the point P $\left(1,3\right)$ and $\left(\frac{-8}{5},\frac{21}{10}\right)$
The slope of the line joining the points $P\left(1,3\right)$ and $Q\left(\frac{-8}{5},\frac{21}{10}\right)$ is given by $m=\frac{\frac{21}{10}-3}{\frac{-8}{5}-1}=\frac{9}{26}$
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
$y =\frac{9}{26}x+3-\left(\frac{9}{26}\right)\left(1\right)$
or, $y =\frac{9}{26}x+3-\frac{9}{26}$
or, $y=\frac{9}{26}x+\frac{69}{26}$