Find the equation of the lines passing through the point of intersection of :
`5x-3y=1` and `2x+3y=23` and perpendicular to the line whose equation is :
`(i) x-2y=3` `(ii) y=0`
`(iii)x=0` `(iv) 5x-3y=1`.

Answer

To find the equation of the lines passing through the point of intersection of the lines \(5x - 3y = 1\) and \(2x + 3y = 23\), and perpendicular to the given lines, we will follow these steps:

### Step 1: Find the point of intersection of the lines
We need to solve the system of equations:
1. \(5x - 3y = 1\) (Equation 1)
2. \(2x + 3y = 23\) (Equation 2)

To eliminate \(y\), we can add both equations. First, we can multiply Equation 1 by 1 and Equation 2 by 1 to align the coefficients of \(y\):

From Equation 1:
\[ 5x - 3y = 1 \]

From Equation 2:
\[ 2x + 3y = 23 \]

Now, add both equations:
\[ (5x - 3y) + (2x + 3y) = 1 + 23 \]
This simplifies to:
\[ 7x = 24 \implies x = \frac{24}{7} \]

Now substitute \(x\) back into one of the original equations to find \(y\). Let's use Equation 1:
\[ 5\left(\frac{24}{7}\right) - 3y = 1 \]
\[ \frac{120}{7} - 3y = 1 \]
\[ -3y = 1 - \frac{120}{7} \]
Convert \(1\) to a fraction with a denominator of \(7\):
\[ -3y = \frac{7}{7} - \frac{120}{7} = \frac{-113}{7} \]
\[ y = \frac{113}{21} \]

Thus, the point of intersection \(P\) is:
\[ P\left(\frac{24}{7}, \frac{113}{21}\right) \]

### Step 2: Find the slope of the given lines
Now we will find the slopes of the lines to which we need to find perpendicular lines.

#### (i) For the line \(x - 2y = 3\):
Rearranging gives:
\[ 2y = x - 3 \implies y = \frac{1}{2}x - \frac{3}{2} \]
The slope \(m_1 = \frac{1}{2}\).

The slope of the perpendicular line \(m\) is given by:
\[ m \cdot m_1 = -1 \implies m \cdot \frac{1}{2} = -1 \implies m = -2 \]

#### (ii) For the line \(y = 0\):
This is the x-axis, which has a slope of \(0\). The slope of the perpendicular line is undefined (vertical line).

#### (iii) For the line \(x = 0\):
This is the y-axis, which has an undefined slope. The slope of the perpendicular line is \(0\) (horizontal line).

#### (iv) For the line \(5x - 3y = 1\):
Rearranging gives:
\[ 3y = 5x - 1 \implies y = \frac{5}{3}x - \frac{1}{3} \]
The slope \(m_2 = \frac{5}{3}\).

The slope of the perpendicular line \(m\) is given by:
\[ m \cdot m_2 = -1 \implies m \cdot \frac{5}{3} = -1 \implies m = -\frac{3}{5} \]

### Step 3: Write the equations of the perpendicular lines
Using the point-slope form \(y - y_1 = m(x - x_1)\):

1. **For the line perpendicular to \(x - 2y = 3\)**:
\[ y - \frac{113}{21} = -2\left(x - \frac{24}{7}\right) \]
Simplifying gives:
\[ y - \frac{113}{21} = -2x + \frac{48}{7} \]
Converting \(\frac{48}{7}\) to a denominator of \(21\):
\[ y = -2x + \frac{144}{21} + \frac{113}{21} = -2x + \frac{257}{21} \]

2. **For the line perpendicular to \(y = 0\)** (vertical line):
This line will be a vertical line through \(P\):
\[ x = \frac{24}{7} \]

3. **For the line perpendicular to \(x = 0\)** (horizontal line):
This line will be a horizontal line through \(P\):
\[ y = \frac{113}{21} \]

4. **For the line perpendicular to \(5x - 3y = 1\)**:
\[ y - \frac{113}{21} = -\frac{3}{5}\left(x - \frac{24}{7}\right) \]
Simplifying gives:
\[ y - \frac{113}{21} = -\frac{3}{5}x + \frac{72}{35} \]
Converting \(\frac{72}{35}\) to a denominator of \(21\):
\[ y = -\frac{3}{5}x + \frac{72}{35} + \frac{113}{21} \]

### Final equations:
1. \(y = -2x + \frac{257}{21}\)
2. \(x = \frac{24}{7}\)
3. \(y = \frac{113}{21}\)
4. \(y = -\frac{3}{5}x + \text{(constant)}\)