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: Name the vertices.
Let \[ A=(-2,2),\quad B=(2,-2),\quad C=(1,1) \] We use the distance formula to find the side lengths.

Step 2: Find the length \(AB\).
Using \[ AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \] we get \[ AB = \sqrt{(2+2)^2+(-2-2)^2} \] \[ = \sqrt{4^2+(-4)^2} \] \[ = \sqrt{16+16} \] \[ = \sqrt{32} \] \[ =4\sqrt{2} \]

Step 3: Find the length \(AC\).
\[ AC = \sqrt{(1+2)^2+(1-2)^2} \] \[ = \sqrt{3^2+(-1)^2} \] \[ = \sqrt{9+1} \] \[ = \sqrt{10} \]

Step 4: Find the length \(BC\).
\[ BC = \sqrt{(1-2)^2+(1+2)^2} \] \[ = \sqrt{(-1)^2+3^2} \] \[ = \sqrt{1+9} \] \[ = \sqrt{10} \]

Step 5: Compare the side lengths.
We obtained \[ AB=4\sqrt{2} \] and \[ AC=BC=\sqrt{10} \] Since two sides are equal, \[ AC=BC, \] the triangle is isosceles.

Step 6: Check whether it is right angled.
Now, \[ AB^2=32 \] and \[ AC^2+BC^2=10+10=20 \] Since \[ 32\neq 20, \] the triangle is not right angled.

Step 7: Final conclusion.
Therefore, the triangle is an \[ \boxed{\text{Isosceles triangle}} \]