CONNECTING CONCEPTS  The lines  $y_1=2x-6$ ,   $y_2=-3x+4$ , and   $y_3=-\frac{1}{2}x+4$   intersect to form the sides of a right triangle. Find the perimeter, to the nearest hundredth and area of the triangle.
Perimeter:  units

Area:  square units

To find the perimeter and area of the triangle formed by the lines $ y_1 = 2x - 6 $, $ y_2 = -3x + 4 $, and $ y_3 = -\frac{1}{2}x + 4 $, we'll need to do the following steps:

1. **Find the points of intersection** of the three lines.
2. **Calculate the lengths of the sides** of the triangle.
3. **Determine if it's a right triangle** by checking if the Pythagorean theorem holds.
4. **Compute the perimeter** and **area** of the triangle.

### Step 1: Find the points of intersection
Let's find the points of intersection of each pair of lines.

- **Intersection of $ y_1 = 2x - 6 $ and $ y_2 = -3x + 4 $**:
  \[
  2x - 6 = -3x + 4
  \]
  \[
  5x = 10 \quad \Rightarrow \quad x = 2
  \]
  Substitute $ x = 2 $ into either equation to find $ y $:
  \[
  y = 2(2) - 6 = 4 - 6 = -2
  \]
  So, the intersection point is $ (2, -2) $.

- **Intersection of $ y_1 = 2x - 6 $ and $ y_3 = -\frac{1}{2}x + 4 $**:
  \[
  2x - 6 = -\frac{1}{2}x + 4
  \]
  \[
  \frac{5x}{2} = 10 \quad \Rightarrow \quad x = 4
  \]
  Substitute $ x = 4 $ into either equation to find $ y $:
  \[
  y = 2(4) - 6 = 8 - 6 = 2
  \]
  So, the intersection point is $ (4, 2) $.

- **Intersection of $ y_2 = -3x + 4 $ and $ y_3 = -\frac{1}{2}x + 4 $**:
  \[
  -3x + 4 = -\frac{1}{2}x + 4
  \]
  \[
  -\frac{5x}{2} = 0 \quad \Rightarrow \quad x = 0
  \]
  Substitute $ x = 0 $ into either equation to find $ y $:
  \[
  y = -\frac{1}{2}(0) + 4 = 4
  \]
  So, the intersection point is $ (0, 4) $.

Thus, the vertices of the triangle are $ (2, -2) $, $ (4, 2) $, and $ (0, 4) $.

### Step 2: Calculate the lengths of the sides
We calculate the distances between each pair of points:

- Length of side $ AB $ between $ (2, -2) $ and $ (4, 2) $:
  \[
  AB = \sqrt{(4 - 2)^2 + (2 - (-2))^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47
  \]

- Length of side $ BC $ between $ (4, 2) $ and $ (0, 4) $:
  \[
  BC = \sqrt{(4 - 0)^2 + (2 - 4)^2} = \sqrt{16 + 4} = \sqrt{20} \approx 4.47
  \]

- Length of side $ CA $ between $ (0, 4) $ and $ (2, -2) $:
  \[
  CA = \sqrt{(2 - 0)^2 + (-2 - 4)^2} = \sqrt{4 + 36} = \sqrt{40} \approx 6.32
  \]

### Step 3: Determine if it's a right triangle
To verify if the triangle is a right triangle, check if the Pythagorean theorem holds:
\[
AB^2 + BC^2 \stackrel{?}{=} CA^2
\]
\[
(4.47)^2 + (4.47)^2 \approx 20 + 20 = 40
\]
\[
CA^2 = 6.32^2 \approx 40
\]
Since the sums match, the triangle is indeed a right triangle.

### Step 4: Compute the perimeter and area
- **Perimeter**:
  \[
  \text{Perimeter} = AB + BC + CA \approx 4.47 + 4.47 + 6.32 = 15.26 \text{ units}
  \]

- **Area**:
  Since it's a right triangle, the area can be calculated using the formula:
  \[
  \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
  \]
  Here, the base and height are $ AB $ and $ BC $ (or vice versa):
  \[
  \text{Area} = \frac{1}{2} \times 4.47 \times 4.47 \approx 10.00 \text{ square units}
  \]

### Final Answers:
- **Perimeter**: $ \approx 15.26 $ units
- **Area**: $ \approx 10.00 $ square units

Would you like more details or have any questions?

Here are some related questions:
1. What is the significance of verifying the triangle as a right triangle?
2. How would the solution change if the triangle were not right?
3. Can you determine the centroid of this triangle?
4. What would happen if one of the lines were parallel to the x-axis?
5. How can you find the circumcenter of this triangle?

**Tip**: In coordinate geometry, always verify the type of triangle using distance formulas and check for special properties like being right-angled.

Learn how to find the perimeter and area of a right triangle formed by the intersection of lines y1 = 2x - 6, y2 = -3x + 4, and y3 = -1/2x + 4. Step-by-step solution includes finding intersection points, verifying it as a right triangle using the Pythagorean theorem, and calculating the perimeter and area. Suitable for Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation