The distance of the line `2x+3y-5=0` from the point (3, 5) along the line `5x-3y=0` in units is

Options

A. `(2sqrt(34))/(21)`
B. `(16)/(21)`
C. `(4sqrt(34))/(21)`
D. `(3sqrt(43))/(21)`

Answer

To find the distance of the line \(2x + 3y - 5 = 0\) from the point \( (3, 5) \) along the line \(5x - 3y = 0\), we can follow these steps:

### Step 1: Identify the slope of the line \(5x - 3y = 0\)
First, we rewrite the equation \(5x - 3y = 0\) in slope-intercept form \(y = mx + c\):
\[ 3y = 5x \implies y = \frac{5}{3}x \]
Thus, the slope \(m\) of the line is \(\frac{5}{3}\).

**Hint:** To find the slope of a line from its equation, rearrange it into the form \(y = mx + c\).

### Step 2: Write the equation of the line passing through the point (3, 5) with the same slope
Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\):
\[ y - 5 = \frac{5}{3}(x - 3) \]
Expanding this:
\[ y - 5 = \frac{5}{3}x - 5 \implies y = \frac{5}{3}x \]
This shows that the line through (3, 5) with slope \(\frac{5}{3}\) is the same as \(5x - 3y = 0\).

**Hint:** The point-slope form is useful for finding the equation of a line given a point and a slope.

### Step 3: Find the intersection of the lines \(2x + 3y - 5 = 0\) and \(5x - 3y = 0\)
We will solve the system of equations:
1. \(2x + 3y - 5 = 0\)
2. \(5x - 3y = 0\)

From the second equation, we can express \(y\) in terms of \(x\):
\[ 3y = 5x \implies y = \frac{5}{3}x \]
Substituting \(y\) into the first equation:
\[ 2x + 3\left(\frac{5}{3}x\right) - 5 = 0 \]
This simplifies to:
\[ 2x + 5x - 5 = 0 \implies 7x - 5 = 0 \implies x = \frac{5}{7} \]
Now substituting \(x\) back to find \(y\):
\[ y = \frac{5}{3}\left(\frac{5}{7}\right) = \frac{25}{21} \]
Thus, the intersection point \(A\) is \(\left(\frac{5}{7}, \frac{25}{21}\right)\).

**Hint:** To find the intersection of two lines, substitute one equation into the other.

### Step 4: Calculate the distance between the points \(A\left(\frac{5}{7}, \frac{25}{21}\right)\) and \(P(3, 5)\)
Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{\left(3 - \frac{5}{7}\right)^2 + \left(5 - \frac{25}{21}\right)^2} \]
Calculating \(3 - \frac{5}{7}\):
\[ 3 = \frac{21}{7} \implies 3 - \frac{5}{7} = \frac{21 - 5}{7} = \frac{16}{7} \]
Calculating \(5 - \frac{25}{21}\):
\[ 5 = \frac{105}{21} \implies 5 - \frac{25}{21} = \frac{105 - 25}{21} = \frac{80}{21} \]
Now substituting back into the distance formula:
\[ d = \sqrt{\left(\frac{16}{7}\right)^2 + \left(\frac{80}{21}\right)^2} \]
Calculating the squares:
\[ d = \sqrt{\frac{256}{49} + \frac{6400}{441}} \]
Finding a common denominator (which is \(441\)):
\[ \frac{256}{49} = \frac{256 \times 9}{441} = \frac{2304}{441} \]
Thus,
\[ d = \sqrt{\frac{2304 + 6400}{441}} = \sqrt{\frac{8704}{441}} = \frac{\sqrt{8704}}{21} \]
Simplifying \(\sqrt{8704}\):
\[ 8704 = 16 \times 34 \implies \sqrt{8704} = 4\sqrt{34} \]
Finally, the distance is:
\[ d = \frac{4\sqrt{34}}{21} \]

### Final Answer:
The distance of the line \(2x + 3y - 5 = 0\) from the point \( (3, 5) \) along the line \(5x - 3y = 0\) is \(\frac{4\sqrt{34}}{21}\) units.

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