The equation of the lines through the point of intersectiono the lines `x-3y+1=0, 2x+5y-9=0` and whose disance from the origin is `sqrt(5)` are

Options

A. `3x+2y-7=0`
`5x-7y+12=0`
B. `2x+y-7=0`
`3x-5y+2=0`
C. `2x+y-5=0`
D. none

Answer

To solve the problem step by step, we need to find the equation of the lines that pass through the point of intersection of the lines \( x - 3y + 1 = 0 \) and \( 2x + 5y - 9 = 0 \), and whose distance from the origin is \( \sqrt{5} \).

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

To find the point of intersection, we can solve the equations simultaneously.

1. The equations are:
\[ x - 3y + 1 = 0 \quad (1) \]
\[ 2x + 5y - 9 = 0 \quad (2) \]

2. From equation (1), we can express \( x \) in terms of \( y \):
\[ x = 3y - 1 \]

3. Substitute this expression for \( x \) into equation (2):
\[ 2(3y - 1) + 5y - 9 = 0 \]
\[ 6y - 2 + 5y - 9 = 0 \]
\[ 11y - 11 = 0 \]
\[ y = 1 \]

4. Substitute \( y = 1 \) back into the expression for \( x \):
\[ x = 3(1) - 1 = 2 \]

So, the point of intersection is \( (2, 1) \).

### Step 2: Find the equation of the line passing through the point of intersection

The general form of the equation of a line can be written as:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) \) is the point of intersection and \( m \) is the slope.

However, we can also express the line in the form:
\[ x - 3y + 1 + \lambda (2x + 5y - 9) = 0 \]
This represents a family of lines passing through the intersection point.

### Step 3: Use the distance formula

The distance \( d \) from the origin \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line, we can express it as:
\[ (1 + 2\lambda)x + (-3 + 5\lambda)y + (1 - 9\lambda) = 0 \]
Here, \( A = 1 + 2\lambda \), \( B = -3 + 5\lambda \), and \( C = 1 - 9\lambda \).

Setting the distance from the origin to \( \sqrt{5} \):
\[ \frac{|1 + 2\lambda(0) + (-3 + 5\lambda)(0) + (1 - 9\lambda)|}{\sqrt{(1 + 2\lambda)^2 + (-3 + 5\lambda)^2}} = \sqrt{5} \]

### Step 4: Solve for \( \lambda \)

1. The numerator simplifies to \( |1 - 9\lambda| \).
2. The denominator simplifies to:
\[ \sqrt{(1 + 2\lambda)^2 + (-3 + 5\lambda)^2} \]
Expanding this gives:
\[ (1 + 4\lambda + 4\lambda^2) + (9 - 30\lambda + 25\lambda^2) = 34\lambda^2 - 26\lambda + 10 \]

3. Setting the equation:
\[ \frac{|1 - 9\lambda|}{\sqrt{34\lambda^2 - 26\lambda + 10}} = \sqrt{5} \]

4. Squaring both sides and solving the resulting quadratic equation will yield values for \( \lambda \).

### Step 5: Find the required lines

After solving for \( \lambda \), substitute back into the equation of the line to get the required lines.