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$$Mathematics

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