The equations of straight line L1 and L2 are x+2y+8=0 and 3x+6y-12=0 respectively. Let P be a moving point in the retangular coordinate plane such that the perpendicular distance from P to L1 is equal to the perpendicular distance from P to L2. Find the equation of the locus of P

To find the equation of the locus of the point \( P(x, y) \) such that the perpendicular distance from \( P \) to the line \( L_1 \) is equal to the perpendicular distance from \( P \) to the line \( L_2 \), we can follow these steps:

### Step 1: Distance from a Point to a Line Formula
The formula for the perpendicular distance from a point \( P(x_1, y_1) \) to a line given by \( Ax + By + C = 0 \) is:

\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]

### Step 2: Equation of Lines
For line \( L_1 \): \( x + 2y + 8 = 0 \)
- Coefficients: \( A_1 = 1 \), \( B_1 = 2 \), \( C_1 = 8 \)

For line \( L_2 \): \( 3x + 6y - 12 = 0 \)
- Coefficients: \( A_2 = 3 \), \( B_2 = 6 \), \( C_2 = -12 \)

### Step 3: Apply Distance Formula
For \( P(x, y) \), the perpendicular distance from \( P \) to \( L_1 \) is:

\[
d_1 = \frac{|x + 2y + 8|}{\sqrt{1^2 + 2^2}} = \frac{|x + 2y + 8|}{\sqrt{5}}
\]

The perpendicular distance from \( P \) to \( L_2 \) is:

\[
d_2 = \frac{|3x + 6y - 12|}{\sqrt{3^2 + 6^2}} = \frac{|3x + 6y - 12|}{\sqrt{45}} = \frac{|3x + 6y - 12|}{3\sqrt{5}}
\]

### Step 4: Set the Distances Equal
Since the distances from \( P \) to both lines are equal, we set \( d_1 = d_2 \):

\[
\frac{|x + 2y + 8|}{\sqrt{5}} = \frac{|3x + 6y - 12|}{3\sqrt{5}}
\]

### Step 5: Simplify the Equation
Canceling \( \sqrt{5} \) from both sides:

\[
|x + 2y + 8| = \frac{1}{3}|3x + 6y - 12|
\]

Multiply both sides by 3:

\[
3|x + 2y + 8| = |3x + 6y - 12|
\]

### Step 6: Solve the Absolute Value Equation
This equation implies two possible cases:

#### Case 1:
\[
3(x + 2y + 8) = 3x + 6y - 12
\]

Simplifying this:

\[
3x + 6y + 24 = 3x + 6y - 12
\]

Canceling terms:

\[
24 = -12
\]

This is a contradiction, so this case gives no solution.

#### Case 2:
\[
3(x + 2y + 8) = -(3x + 6y - 12)
\]

Simplifying this:

\[
3x + 6y + 24 = -3x - 6y + 12
\]

Bringing all terms to one side:

\[
6x + 12y + 12 = 0
\]

Dividing the entire equation by 6:

\[
x + 2y + 2 = 0
\]

### Step 7: Final Answer
The equation of the locus of point \( P \) is:

\[
x + 2y + 2 = 0
\]

Would you like more details or have any questions? Here are 5 related questions:

1. How do you find the perpendicular distance from a point to a line?
2. What are the conditions for two lines to be parallel or perpendicular?
3. Can the distance formula be applied in 3D space as well?
4. How would the equation change if the lines were not parallel?
5. How do absolute value equations affect geometrical interpretations?

**Tip**: When solving absolute value equations, always consider both the positive and negative cases to cover all possibilities.

The equations of straight line L1 and L2 are x+2y+8=0 and 3x+6y-12=0 respectively. Let P be a moving point in the rectangular coordinate plane such that the perpendicular distance from P to L1 is equal to the perpendicular distance from P to L2. Find the equation of the locus of P.

Learn how to find the equation of the locus of a point P that is equidistant from two given lines in a rectangular coordinate plane. This problem involves key concepts from coordinate geometry, such as the perpendicular distance formula and absolute value equations. Suitable for students in Grades 10-12.

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