Let L be the line passing through the point P (1,2) such that its intercepted segment between the co-ordinate axes is bisected at P. If `L_(1)` is the line perpendicular to L and psssing through the point (-2,1), then the point of intersection of L and `L_(1)` is

Options

A. `(4/5,12/5)`
B. `(3/5,23/10)`
C. `(11/20),(29/10)`
D. `(3/10,17/5)`

Answer

To solve the problem step by step, we will find the equations of the lines \( L \) and \( L_1 \) and then determine their point of intersection.

### Step 1: Determine the intercepts of line \( L \)

Given that the line \( L \) passes through the point \( P(1, 2) \) and its intercepted segment between the coordinate axes is bisected at \( P \), we can denote the x-intercept as \( (x_1, 0) \) and the y-intercept as \( (0, y_1) \).

Since \( P \) is the midpoint of the segment between the intercepts, we have:
\[ \frac{x_1 + 0}{2} = 1 \quad \text{and} \quad \frac{0 + y_1}{2} = 2 \]

### Step 2: Solve for the intercepts

From the first equation:
\[ \frac{x_1}{2} = 1 \implies x_1 = 2 \]
From the second equation:
\[ \frac{y_1}{2} = 2 \implies y_1 = 4 \]

Thus, the intercepts are \( (2, 0) \) and \( (0, 4) \).

### Step 3: Find the equation of line \( L \)

Using the intercept form of the line equation:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
where \( a = 2 \) and \( b = 4 \), we have:
\[ \frac{x}{2} + \frac{y}{4} = 1 \]

Multiplying through by 4 to eliminate the denominators:
\[ 2x + y = 4 \]
or rearranging gives:
\[ 2x + y - 4 = 0 \]

### Step 4: Find the slope of line \( L \)

The slope \( m_1 \) of line \( L \) can be found from the equation \( y = -2x + 4 \), which gives:
\[ m_1 = -2 \]

### Step 5: Find the slope of line \( L_1 \)

Since \( L_1 \) is perpendicular to \( L \), the slope \( m_2 \) of line \( L_1 \) is given by:
\[ m_1 \cdot m_2 = -1 \implies -2 \cdot m_2 = -1 \implies m_2 = \frac{1}{2} \]

### Step 6: Write the equation of line \( L_1 \)

Line \( L_1 \) passes through the point \( (-2, 1) \). Using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
we substitute \( (x_1, y_1) = (-2, 1) \) and \( m = \frac{1}{2} \):
\[ y - 1 = \frac{1}{2}(x + 2) \]
Simplifying this:
\[ y - 1 = \frac{1}{2}x + 1 \implies y = \frac{1}{2}x + 2 \]

### Step 7: Find the point of intersection of lines \( L \) and \( L_1 \)

We have the equations:
1. \( 2x + y - 4 = 0 \) (line \( L \))
2. \( y = \frac{1}{2}x + 2 \) (line \( L_1 \))

Substituting the second equation into the first:
\[ 2x + \left(\frac{1}{2}x + 2\right) - 4 = 0 \]
Simplifying:
\[ 2x + \frac{1}{2}x + 2 - 4 = 0 \implies 2x + \frac{1}{2}x - 2 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 4x + x - 4 = 0 \implies 5x - 4 = 0 \implies x = \frac{4}{5} \]

Now substituting \( x = \frac{4}{5} \) back into the equation of line \( L_1 \):
\[ y = \frac{1}{2}\left(\frac{4}{5}\right) + 2 = \frac{2}{5} + 2 = \frac{2}{5} + \frac{10}{5} = \frac{12}{5} \]

### Final Result

The point of intersection of lines \( L \) and \( L_1 \) is:
\[ \left(\frac{4}{5}, \frac{12}{5}\right) \]