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

Options

A. `(130)/(17sqrt129)`
B. `(13)/(7sqrt29)`
C. `(130)/(7)`
D. None of these

Explanation

To solve the problem of finding the distance from the point of intersection of the lines \(2x - 3y + 5 = 0\) and \(3x + 4y = 0\) to the line \(5x - 2y = 0\), we will follow these steps:

### Step 1: Find the point of intersection of the lines

We have two equations:
1. \(2x - 3y + 5 = 0\)
2. \(3x + 4y = 0\)

To find the intersection, we can solve these equations simultaneously.

From the second equation, we can express \(y\) in terms of \(x\):
\[ 4y = -3x \implies y = -\frac{3}{4}x \]

Now, substitute \(y\) in the first equation:
\[ 2x - 3\left(-\frac{3}{4}x\right) + 5 = 0 \]
\[ 2x + \frac{9}{4}x + 5 = 0 \]
To eliminate the fraction, multiply through by 4:
\[ 8x + 9x + 20 = 0 \implies 17x + 20 = 0 \implies x = -\frac{20}{17} \]

Now substitute \(x\) back to find \(y\):
\[ y = -\frac{3}{4}\left(-\frac{20}{17}\right) = \frac{60}{68} = \frac{15}{17} \]

Thus, the point of intersection is:
\[ \left(-\frac{20}{17}, \frac{15}{17}\right) \]

### Step 2: Find the distance from the point to the line

The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

For the line \(5x - 2y = 0\), we can rewrite it as:
\[ 5x - 2y + 0 = 0 \]
Here, \(A = 5\), \(B = -2\), and \(C = 0\).

Now substituting the point \(\left(-\frac{20}{17}, \frac{15}{17}\right)\) into the distance formula:
\[ d = \frac{|5\left(-\frac{20}{17}\right) - 2\left(\frac{15}{17}\right) + 0|}{\sqrt{5^2 + (-2)^2}} \]
Calculating the numerator:
\[ = \frac{|- \frac{100}{17} - \frac{30}{17}|}{\sqrt{25 + 4}} = \frac{|- \frac{130}{17}|}{\sqrt{29}} = \frac{\frac{130}{17}}{\sqrt{29}} \]

Thus, the distance is:
\[ d = \frac{130}{17\sqrt{29}} \]

### Final Answer
The distance of the point of intersection from the line \(5x - 2y = 0\) is:
\[ \frac{130}{17\sqrt{29}} \]