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Q.
The distance of the point of intersection of the lines $2x- 3y + 5 = 0$ and $3x + 4y=0$ from the line $5x- 2y = 0$ is
Straight Lines
Solution:
On solving the equations $2x - 3y + 5 = 0$ and $3x + 4y = 0$,
we get $x=\frac{-20}{17}$, $y=\frac{15}{17}$
$\therefore $ Point of intersection is $\left(\frac{-20}{17}, \frac{15}{17}\right)$
$\therefore $ Distance $=\frac{\left|5\left(\frac{-20}{17}\right)-2\left(\frac{15}{17}\right)\right|}{\sqrt{25+4}}$
$=\frac{130}{17\sqrt{29}}$