Find the equation of line passing through the point of intersection of the lines `2x+3y+1=0` and `3x-5y-5=0`
`(i)` perpendicular to `X`-axis.
`(ii)` perpendicular to `Y`-axis.
`(iii)` perpendicular to line `x-2y+1=0`
`(iv)` parallel to line `x+2y-1=0`.

Answer

To solve the problem, we need to find the equations of lines that pass through the point of intersection of the lines \(2x + 3y + 1 = 0\) and \(3x - 5y - 5 = 0\). We will find the point of intersection first and then derive the required equations based on the conditions given.

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

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

We can rewrite these equations as:
1. \(2x + 3y = -1\)
2. \(3x - 5y = 5\)

To solve these equations, we can use the method of elimination. Let's multiply Equation 1 by 5 and Equation 2 by 3 to eliminate \(y\):

\[ 5(2x + 3y) = 5(-1) \implies 10x + 15y = -5 \quad \text{(Equation 3)} \]
\[ 3(3x - 5y) = 3(5) \implies 9x - 15y = 15 \quad \text{(Equation 4)} \]

Now, add Equation 3 and Equation 4:

\[ 10x + 15y + 9x - 15y = -5 + 15 \]
\[ 19x = 10 \implies x = \frac{10}{19} \]

Now substitute \(x = \frac{10}{19}\) back into Equation 1 to find \(y\):

\[ 2\left(\frac{10}{19}\right) + 3y = -1 \]
\[ \frac{20}{19} + 3y = -1 \implies 3y = -1 - \frac{20}{19} = -\frac{19}{19} - \frac{20}{19} = -\frac{39}{19} \]
\[ y = -\frac{39}{57} = -\frac{13}{19} \]

Thus, the point of intersection is:

\[ P\left(\frac{10}{19}, -\frac{13}{19}\right) \]

### Step 2: Find the equations of the required lines

#### (i) Line perpendicular to the X-axis

A line perpendicular to the X-axis is a vertical line. The equation of such a line passing through the point \(P\) is given by:

\[ x = \frac{10}{19} \]

#### (ii) Line perpendicular to the Y-axis

A line perpendicular to the Y-axis is a horizontal line. The equation of such a line passing through the point \(P\) is given by:

\[ y = -\frac{13}{19} \]

#### (iii) Line perpendicular to the line \(x - 2y + 1 = 0\)

First, we need to find the slope of the line \(x - 2y + 1 = 0\). Rearranging gives:

\[ 2y = x + 1 \implies y = \frac{1}{2}x + \frac{1}{2} \]

The slope (m) of this line is \(\frac{1}{2}\). The slope of the line perpendicular to it will be the negative reciprocal:

\[ m_{\text{perpendicular}} = -\frac{1}{\frac{1}{2}} = -2 \]

Using the point-slope form of the equation of a line:

\[ y - y_1 = m(x - x_1) \]

Substituting \(P\left(\frac{10}{19}, -\frac{13}{19}\right)\):

\[ y + \frac{13}{19} = -2\left(x - \frac{10}{19}\right) \]

Simplifying:

\[ y + \frac{13}{19} = -2x + \frac{20}{19} \]
\[ y = -2x + \frac{20}{19} - \frac{13}{19} \]
\[ y = -2x + \frac{7}{19} \]

#### (iv) Line parallel to the line \(x + 2y - 1 = 0\)

The slope of the line \(x + 2y - 1 = 0\) can be found by rearranging:

\[ 2y = -x + 1 \implies y = -\frac{1}{2}x + \frac{1}{2} \]

The slope is \(-\frac{1}{2}\). A line parallel to this will have the same slope. Using the point-slope form:

\[ y - y_1 = m(x - x_1) \]

Substituting \(P\left(\frac{10}{19}, -\frac{13}{19}\right)\):

\[ y + \frac{13}{19} = -\frac{1}{2}\left(x - \frac{10}{19}\right) \]

Simplifying:

\[ y + \frac{13}{19} = -\frac{1}{2}x + \frac{5}{19} \]
\[ y = -\frac{1}{2}x + \frac{5}{19} - \frac{13}{19} \]
\[ y = -\frac{1}{2}x - \frac{8}{19} \]

### Summary of Results

1. Perpendicular to X-axis: \(x = \frac{10}{19}\)
2. Perpendicular to Y-axis: \(y = -\frac{13}{19}\)
3. Perpendicular to \(x - 2y + 1 = 0\): \(y = -2x + \frac{7}{19}\)
4. Parallel to \(x + 2y - 1 = 0\): \(y = -\frac{1}{2}x - \frac{8}{19}\)