The distance of the point of intersection

A. $\frac{130}{17 \sqrt{29}}$

Explanation:

Given two lines are 2x – 3y + 5 = 0 … (i)

And 3x + 4y = 0 … (ii)

Now, point of intersection of these lines can be find out as

Multiplying equation (i) by 3, we get

6x – 9y + 15 = 0 … (iii)

Multiplying equation (ii) by 2, we get

6x + 8y = 0 … (iv)

On subtracting equation (iv) from (iii), we get

6x – 9y + 15 – 6x – 8y = 0

⇒ – 17y + 15 = 0

⇒ – 17y = -15

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

On putting value of $y$ in equation (ii), we get

$3 x+4\left(\frac{15}{17}\right)=0$

$\Rightarrow 3 x=-\frac{60}{17}$

$\Rightarrow \mathrm{x}=-\frac{20}{17}$

So, the point of intersection of given two lines is

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

Now, perpendicular distance from the point $\left(-\frac{20}{17}, \frac{15}{17}\right)$ to the given line $5 x-2 y$ $=0$

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

$\Rightarrow \mathrm{d}=\left|\frac{-\frac{100}{17}-\frac{30}{17}}{\sqrt{25+4}}\right|$

$\Rightarrow \mathrm{d}=\frac{130}{17 \sqrt{29}}$

Hence, the correct option is (a)