Key Concepts and Formulas

• Intersection of Lines: To find the point of intersection of two lines, solve their equations simultaneously.
• Distance Formula: The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• Slope of a Line: The slope $m$ of a line $ax + by + c = 0$ is given by $m = -\frac{a}{b}$. Two lines are perpendicular if the product of their slopes is $-1$.

Step-by-Step Solution

Step 1: Find the intersection points of the lines.

We need to find the vertices of the triangle formed by the intersection of the three lines:

1. $x - y = 0$
2. $x + 2y = 3$
3. $2x + y = 6$

• Intersection of lines 1 and 2: Substituting $x = y$ from equation (1) into equation (2), we get: $y + 2y = 3 \implies 3y = 3 \implies y = 1$. Since $x = y$, we have $x = 1$. Thus, the intersection point is $(1, 1)$.

• Intersection of lines 1 and 3: Substituting $x = y$ from equation (1) into equation (3), we get: $2y + y = 6 \implies 3y = 6 \implies y = 2$. Since $x = y$, we have $x = 2$. Thus, the intersection point is $(2, 2)$.

• Intersection of lines 2 and 3: We have the system of equations: $x + 2y = 3$ $2x + y = 6$ Multiply the first equation by 2 to get $2x + 4y = 6$. Subtract the second equation from this to get: $(2x + 4y) - (2x + y) = 6 - 6 \implies 3y = 0 \implies y = 0$. Substituting $y = 0$ into $x + 2y = 3$, we get $x + 2(0) = 3 \implies x = 3$. Thus, the intersection point is $(3, 0)$.

So the vertices of the triangle are $A(1, 1)$, $B(2, 2)$, and $C(3, 0)$.

Step 2: Calculate the lengths of the sides of the triangle.

We use the distance formula to find the lengths of the sides:

• $AB = \sqrt{(2 - 1)^2 + (2 - 1)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$
• $BC = \sqrt{(3 - 2)^2 + (0 - 2)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$
• $AC = \sqrt{(3 - 1)^2 + (0 - 1)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$

Since $BC = AC$, the triangle is isosceles.

Step 3: Check if the triangle is right-angled.

To check if the triangle is right-angled, we can check if the sides satisfy the Pythagorean theorem. Since $AC = BC = \sqrt{5}$, let's check if $AB^2 + AC^2 = BC^2$, $AB^2 + BC^2 = AC^2$ or $AC^2 + BC^2 = AB^2$

• $AB^2 = 2, BC^2 = 5, AC^2 = 5$. If $AB^2 + BC^2 = AC^2$, then $2 + 5 = 5$ which is false. If $AB^2 + AC^2 = BC^2$, then $2 + 5 = 5$ which is false. If $AC^2 + BC^2 = AB^2$, then $5 + 5 = 2$, which is false.

However, an alternative way to check if it's a right triangle is to calculate the slopes of the sides and see if any two sides are perpendicular (product of slopes is -1).

• Slope of $AB = \frac{2 - 1}{2 - 1} = \frac{1}{1} = 1$
• Slope of $BC = \frac{0 - 2}{3 - 2} = \frac{-2}{1} = -2$
• Slope of $AC = \frac{0 - 1}{3 - 1} = \frac{-1}{2} = -\frac{1}{2}$

Since the slope of $AC \times$ slope of $BC = -2 \times -\frac{1}{2} = 1 \ne -1$ and the slope of $AB \times$ slope of $BC = 1 \times -2 = -2 \ne -1$ and the slope of $AB \times$ slope of $AC = 1 \times -\frac{1}{2} = -\frac{1}{2} \ne -1$, the triangle is not right-angled.

Let's re-examine the problem statement and the given answer. It's possible there's an error in the question or the provided answer. Let's verify if the triangle is indeed right-angled.

The equations of the lines are: $L_1: x - y = 0$ $L_2: x + 2y = 3$ $L_3: 2x + y = 6$

Their slopes are: $m_1 = 1$ $m_2 = -\frac{1}{2}$ $m_3 = -2$

The slopes of the sides of the triangle are: $AB: \frac{2-1}{2-1} = 1$ $BC: \frac{0-2}{3-2} = -2$ $AC: \frac{0-1}{3-1} = -\frac{1}{2}$ $m_{AB} = 1, m_{BC} = -2, m_{AC} = -\frac{1}{2}$ The lines $L_2$ and $L_3$ that intersect at $C(3,0)$ have slopes $-\frac{1}{2}$ and $-2$. The product of these slopes is $1 \ne -1$. The vertices are A(1,1), B(2,2), C(3,0) If we check whether the line $AC$ is perpendicular to $BC$, we have slopes $-\frac{1}{2}$ and $-2$. Since the product of these slopes is 1, the lines are not perpendicular. However, we consider the slopes of the sides of the triangle $m_{AB}, m_{BC}, m_{AC}$. $m_{AB} \cdot m_{AC} = 1 \cdot (-\frac{1}{2}) = -\frac{1}{2} \ne -1$ $m_{AB} \cdot m_{BC} = 1 \cdot (-2) = -2 \ne -1$ $m_{AC} \cdot m_{BC} = (-\frac{1}{2}) \cdot (-2) = 1 \ne -1$.

Since the product of none of the slopes is -1, the triangle is NOT right-angled.

I have confirmed that the triangle is isosceles with sides $\sqrt{5}$, $\sqrt{5}$ and $\sqrt{2}$, and not a right-angled triangle. Thus, the correct answer from the options is (D).

Common Mistakes & Tips

• Double-check the calculations for intersection points and distances to avoid errors.
• Remember that the product of slopes of perpendicular lines is -1.
• When calculating the sides of a triangle, make sure you are using the correct vertices.

Summary

We found the intersection points of the three lines to determine the vertices of the triangle. Then, we calculated the lengths of the sides using the distance formula and found the slopes of the sides. By comparing the side lengths, we determined that the triangle is isosceles. Checking the slopes, we found that the triangle is not right-angled, equilateral, or scalene.

The final answer is \boxed{D}.