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Question

Find the equation of line through the intersection of $$5x-3y=1$$ and $$2x+3y=23$$ and perpendicular to the line $$5x-3y-1=0$$


Solution

Point of intersection of $$5x - 3y = 1 $$ and $$ 2x+3y=23$$ is $$\left (\dfrac{24}{7},\dfrac{113}{21}\right)$$
And slope of $$5x-3y-1=0$$ is $$\left(\dfrac{5}{3}\right)$$ $$ \qquad (by\ differentiating \ eq. w.r.t. \ x)$$
Then slope of line perpendicular to it is $$\left(-\dfrac{3}{5}\right)$$


$$\therefore \ $$ Eq. of line passing through point $$\left (\dfrac{24}{7},\dfrac{113}{21}\right)$$ and having slope $$\left(-\dfrac{3}{5}\right)$$ is given by $$ \left(y-\dfrac{113}{21}\right)=\left(-\dfrac{3}{5}\right) \left (x-\dfrac{24}{7}\right)$$
By solving we get equation of line $$\Rightarrow 105y+63x=781$$
829825_881351_ans_de427cc949d844d1a61fe9f87b4f686f.jpg

Mathematics

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