Find the equation of the lines passing through the point of intersection of `x+2y=5` and `x-3y=7` and passing through :
`(i) (1,0)` `(ii) (0,-1)`

Answer

To find the equations of the lines passing through the point of intersection of the lines \( x + 2y = 5 \) and \( x - 3y = 7 \), and also passing through the points \( (1,0) \) and \( (0,-1) \), we will follow these steps:

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

We have the equations:
1. \( x + 2y = 5 \) (Equation 1)
2. \( x - 3y = 7 \) (Equation 2)

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

From Equation 1, we can express \( x \) in terms of \( y \):
\[ x = 5 - 2y \]

Now, substitute this expression for \( x \) into Equation 2:
\[ (5 - 2y) - 3y = 7 \]
Simplifying this gives:
\[ 5 - 5y = 7 \]
\[ -5y = 7 - 5 \]
\[ -5y = 2 \quad \Rightarrow \quad y = -\frac{2}{5} \]

Now substitute \( y = -\frac{2}{5} \) back into Equation 1 to find \( x \):
\[ x + 2\left(-\frac{2}{5}\right) = 5 \]
\[ x - \frac{4}{5} = 5 \]
\[ x = 5 + \frac{4}{5} = \frac{25}{5} + \frac{4}{5} = \frac{29}{5} \]

Thus, the point of intersection is:
\[ \left( \frac{29}{5}, -\frac{2}{5} \right) \]

### Step 2: Find the equation of the line passing through the point of intersection and point \( (1,0) \).

Using the point-slope form of the line equation, where the slope \( m \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Let \( (x_1, y_1) = \left( \frac{29}{5}, -\frac{2}{5} \right) \) and \( (x_2, y_2) = (1, 0) \):
\[ m = \frac{0 - \left(-\frac{2}{5}\right)}{1 - \frac{29}{5}} = \frac{\frac{2}{5}}{1 - \frac{29}{5}} = \frac{\frac{2}{5}}{\frac{5 - 29}{5}} = \frac{\frac{2}{5}}{-\frac{24}{5}} = -\frac{2}{24} = -\frac{1}{12} \]

Now, using the point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - \left(-\frac{2}{5}\right) = -\frac{1}{12}\left(x - \frac{29}{5}\right) \]
\[ y + \frac{2}{5} = -\frac{1}{12}x + \frac{29}{60} \]
Multiplying through by \( 60 \) to eliminate fractions:
\[ 60y + 24 = -5x + 29 \]
Rearranging gives:
\[ 5x + 60y - 5 = 0 \]
Dividing through by 5:
\[ x + 12y - 1 = 0 \]

### Step 3: Find the equation of the line passing through the point of intersection and point \( (0,-1) \).

Using the same point-slope form:
Let \( (x_1, y_1) = \left( \frac{29}{5}, -\frac{2}{5} \right) \) and \( (x_2, y_2) = (0, -1) \):
\[ m = \frac{-1 - \left(-\frac{2}{5}\right)}{0 - \frac{29}{5}} = \frac{-1 + \frac{2}{5}}{-\frac{29}{5}} = \frac{-\frac{5}{5} + \frac{2}{5}}{-\frac{29}{5}} = \frac{-\frac{3}{5}}{-\frac{29}{5}} = \frac{3}{29} \]

Using the point-slope form:
\[ y - \left(-\frac{2}{5}\right) = \frac{3}{29}\left(x - \frac{29}{5}\right) \]
\[ y + \frac{2}{5} = \frac{3}{29}x - \frac{3 \cdot 29}{29 \cdot 5} \]
Multiplying through by \( 145 \) (LCM of 5 and 29):
\[ 145y + 58 = 15x - 87 \]
Rearranging gives:
\[ 15x - 145y - 29 = 0 \]

### Final Answers:
1. The equation of the line passing through \( (1,0) \) is:
\[ x + 12y - 1 = 0 \]
2. The equation of the line passing through \( (0,-1) \) is:
\[ 15x - 145y - 29 = 0 \]