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.