Find out the perimeter of the triangle formed by the intersection of following 3 lines. L1: 2x+3y-6=0 L2: 3x+2y+6=0 L3: 3x-3y+6=0

To find the perimeter of the triangle formed by the intersection of the three lines \( L_1: 2x + 3y - 6 = 0 \), \( L_2: 3x + 2y + 6 = 0 \), and \( L_3: 3x - 3y + 6 = 0 \), we first need to determine the coordinates of the points of intersection.

### Step 1: Find the points of intersection

1. **Intersection of \( L_1 \) and \( L_2 \):**

   \[
   \begin{cases}
   2x + 3y - 6 = 0 \\
   3x + 2y + 6 = 0
   \end{cases}
   \]

   Multiply the first equation by 3 and the second by 2 to eliminate \( x \):

   \[
   6x + 9y - 18 = 0 \quad \text{(i)}
   \]
   \[
   6x + 4y + 12 = 0 \quad \text{(ii)}
   \]

   Subtract equation (ii) from equation (i):

   \[
   (6x + 9y - 18) - (6x + 4y + 12) = 0
   \]
   \[
   5y - 30 = 0
   \]
   \[
   y = 6
   \]

   Substitute \( y = 6 \) back into \( 2x + 3y - 6 = 0 \):

   \[
   2x + 3(6) - 6 = 0
   \]
   \[
   2x + 18 - 6 = 0
   \]
   \[
   2x + 12 = 0
   \]
   \[
   x = -6
   \]

   So, the intersection point is \( (-6, 6) \).

2. **Intersection of \( L_2 \) and \( L_3 \):**

   \[
   \begin{cases}
   3x + 2y + 6 = 0 \\
   3x - 3y + 6 = 0
   \end{cases}
   \]

   Subtract the second equation from the first:

   \[
   (3x + 2y + 6) - (3x - 3y + 6) = 0
   \]
   \[
   5y = 0
   \]
   \[
   y = 0
   \]

   Substitute \( y = 0 \) back into \( 3x + 2y + 6 = 0 \):

   \[
   3x + 2(0) + 6 = 0
   \]
   \[
   3x + 6 = 0
   \]
   \[
   x = -2
   \]

   So, the intersection point is \( (-2, 0) \).

3. **Intersection of \( L_1 \) and \( L_3 \):**

   \[
   \begin{cases}
   2x + 3y - 6 = 0 \\
   3x - 3y + 6 = 0
   \end{cases}
   \]

   Add the two equations:

   \[
   (2x + 3y - 6) + (3x - 3y + 6) = 0
   \]
   \[
   5x = 0
   \]
   \[
   x = 0
   \]

   Substitute \( x = 0 \) back into \( 2x + 3y - 6 = 0 \):

   \[
   2(0) + 3y - 6 = 0
   \]
   \[
   3y = 6
   \]
   \[
   y = 2
   \]

   So, the intersection point is \( (0, 2) \).

### Step 2: Calculate the distances between the points

1. **Distance between \( (-6, 6) \) and \( (-2, 0) \):**

   \[
   d_1 = \sqrt{((-2) - (-6))^2 + (0 - 6)^2}
   \]
   \[
   d_1 = \sqrt{(4)^2 + (-6)^2}
   \]
   \[
   d_1 = \sqrt{16 + 36}
   \]
   \[
   d_1 = \sqrt{52}
   \]
   \[
   d_1 = 2\sqrt{13}
   \]

2. **Distance between \( (-2, 0) \) and \( (0, 2) \):**

   \[
   d_2 = \sqrt{(0 - (-2))^2 + (2 - 0)^2}
   \]
   \[
   d_2 = \sqrt{(2)^2 + (2)^2}
   \]
   \[
   d_2 = \sqrt{4 + 4}
   \]
   \[
   d_2 = \sqrt{8}
   \]
   \[
   d_2 = 2\sqrt{2}
   \]

3. **Distance between \( (0, 2) \) and \( (-6, 6) \):**

   \[
   d_3 = \sqrt{((-6) - 0)^2 + (6 - 2)^2}
   \]
   \[
   d_3 = \sqrt{(-6)^2 + (4)^2}
   \]
   \[
   d_3 = \sqrt{36 + 16}
   \]
   \[
   d_3 = \sqrt{52}
   \]
   \[
   d_3 = 2\sqrt{13}
   \]

### Step 3: Calculate the perimeter

The perimeter \( P \) of the triangle is the sum of the lengths of its sides:

\[
P = d_1 + d_2 + d_3
\]
\[
P = 2\sqrt{13} + 2\sqrt{2} + 2\sqrt{13}
\]
\[
P = 4\sqrt{13} + 2\sqrt{2}
\]

So, the perimeter of the triangle is \( \boxed{4\sqrt{13} + 2\sqrt{2}} \).

Do you want details or have any questions?

Here are some questions you might want to ask:
1. How do you find the area of a triangle given its vertices?
2. Can you explain how to solve systems of linear equations graphically?
3. What is the significance of the slope in linear equations?
4. How can you determine if three points are collinear?
5. What is the distance formula used in coordinate geometry?

Tip: Remember to double-check your calculations for accuracy, especially when dealing with square roots and distances.

Learn how to find the perimeter of a triangle formed by the intersection of the lines 2x+3y-6=0, 3x+2y+6=0, and 3x-3y+6=0 using coordinate geometry and the distance formula. Detailed step-by-step solution provided.

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