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

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Solution

Point of intersection of $5 x - 3 y = 1$ and $2 x + 3 y = 23$ is $(\frac{24}{7}, \frac{113}{21})$
And slope of $5 x - 3 y - 1 = 0$ is $(\frac{5}{3})$ $(b y d i f f e r e n t i a t i n g e q . w . r . t . x)$
Then slope of line perpendicular to it is $(- \frac{3}{5})$

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

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