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

Options

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

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: Find the equation of line \( L \)

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

Since \( P \) is the midpoint of the intercepts, we have:
- The x-coordinate of \( P \) is the average of the x-intercept and 0:
\[ \frac{A + 0}{2} = 1 \implies A = 2 \]
- The y-coordinate of \( P \) is the average of the y-intercept and 0:
\[ \frac{B + 0}{2} = 2 \implies B = 4 \]

Now, we can write the equation of line \( L \) in intercept form:
\[ \frac{x}{A} + \frac{y}{B} = 1 \implies \frac{x}{2} + \frac{y}{4} = 1 \]

Multiplying through by 4 to eliminate the denominators gives:
\[ 2x + y = 4 \quad \text{(Equation 1)} \]

### Step 2: Find the equation of line \( L_1 \)

Line \( L_1 \) is perpendicular to line \( L \) and passes through the point \( (-2, 1) \).

First, we need to find the slope of line \( L \). The equation \( 2x + y = 4 \) can be rearranged to slope-intercept form:
\[ y = -2x + 4 \]
Thus, the slope \( m_L \) of line \( L \) is \( -2 \).

The slope of line \( L_1 \), which is perpendicular to \( L \), will be the negative reciprocal of \( m_L \):
\[ m_{L_1} = \frac{1}{2} \]

Using the point-slope form of the equation of a line, we can write the equation of line \( L_1 \):
\[ y - 1 = \frac{1}{2}(x + 2) \]
Expanding this gives:
\[ y - 1 = \frac{1}{2}x + 1 \implies y = \frac{1}{2}x + 2 \]

To put this in standard form:
\[ x - 2y + 4 = 0 \quad \text{or} \quad x - 2y = -4 \quad \text{(Equation 2)} \]

### Step 3: Find the intersection of lines \( L \) and \( L_1 \)

Now we will solve the system of equations formed by Equation 1 and Equation 2:
1. \( 2x + y = 4 \)
2. \( x - 2y = -4 \)

We can solve this system using substitution or elimination. Here, we will use substitution.

From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 4 - 2x \]

Substituting this expression for \( y \) into Equation 2:
\[ x - 2(4 - 2x) = -4 \]
Expanding and simplifying:
\[ x - 8 + 4x = -4 \implies 5x - 8 = -4 \implies 5x = 4 \implies x = \frac{4}{5} \]

Now substituting \( x = \frac{4}{5} \) back into the expression for \( y \):
\[ y = 4 - 2\left(\frac{4}{5}\right) = 4 - \frac{8}{5} = \frac{20}{5} - \frac{8}{5} = \frac{12}{5} \]

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

### Final Answer:
The point of intersection of lines \( L \) and \( L_1 \) is \( \left(\frac{4}{5}, \frac{12}{5}\right) \).

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