Question

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

• A. $\frac{130}{17\sqrt{29}}$
• B. $\frac{13}{17\sqrt{29}}$
• C. $\frac{130}{17}$
• D. None of these

Answer 1

The correct option is A $\frac{130}{17\sqrt{29}}$

$2x-3y=-5\times 3$

$3x+4y=0\times 2$

$6x-9y=-156x+8y=0----------------17y=-15$

$y=\frac{15}{17}$

$x=\frac{-4y}{3}=\frac{-4}{3}\left(\frac{15}{17}\right)=\frac{-20}{17}$

Distance $\Rightarrow \left|\frac{5\left(\frac{-20}{17}\right)-2\left(\frac{15}{17}\right)}{\sqrt{5^2+2^2}}\right|$

$\Rightarrow \left|\frac{-100-30}{17\sqrt{29}}|\right|$

$\Rightarrow \frac{130}{17\sqrt{29}}$