Question

The triangle with vertices at \[ (-2,2),\ (2,-2)\ \text{and}\ (1,1) \] is

• A. Scalene triangle
• B. Equilateral triangle
• C. Isosceles triangle
• D. Right-angled triangle

Hint:

To classify a triangle using coordinates, first calculate all side lengths using the distance formula and then compare them.

Answer 1

The Correct Option is C

Step 1: Label the vertices.
Let $A = (-2, 2)$, $B = (2, -2)$, and $C = (1, 1)$. We will use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to find all three side lengths.
Step 2: Compute length $AB$.
\[ AB = \sqrt{(2-(-2))^2 + (-2-2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]
Step 3: Compute length $AC$.
\[ AC = \sqrt{(1-(-2))^2 + (1-2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \]
Step 4: Compute length $BC$.
\[ BC = \sqrt{(1-2)^2 + (1-(-2))^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \]
Step 5: Compare the three sides.
We have $AB = 4\sqrt{2}$, $AC = \sqrt{10}$, $BC = \sqrt{10}$. Since $AC = BC$ but $AB \neq AC$, exactly two sides are equal, so the triangle is isosceles.
Step 6: Check if it is right-angled.
For a right angle at $C$, we would need $AB^2 = AC^2 + BC^2$, i.e. $32 = 10 + 10 = 20$. Since $32 \neq 20$, the triangle is not right-angled.
Step 7: State the final answer.
The triangle is \[ \boxed{\text{Isosceles triangle}} \]