Question

The points \( (1,5), (2,3) \) and \( (-2, -11) \) form a:

• A. triangle
• B. parallelogram
• C. square
• D. They are collinear

Hint:

For verifying collinearity or triangle formation, use the distance formula and compare whether the points lie on a straight line or form a closed shape.

Answer 1

The Correct Option is A

Step 1: Use the distance formula between each pair of points. 
The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Let the points be: \[ A = (1,5), \quad B = (2,3), \quad C = (-2,-11) \] Compute distances: \[ AB = \sqrt{(2-1)^2 + (3-5)^2} = \sqrt{1 + 4} = \sqrt{5} \] \[ BC = \sqrt{(-2-2)^2 + (-11-3)^2} = \sqrt{16 + 196} = \sqrt{212} \] \[ CA = \sqrt{(1+2)^2 + (5+11)^2} = \sqrt{9 + 256} = \sqrt{265} \] 

Step 2: Check for collinearity. 
If the points were collinear, the sum of the lengths of the two shorter sides would equal the third. 
\[ \sqrt{5} + \sqrt{212} \neq \sqrt{265} \] So the points are not collinear. 

Step 3: Check for square or parallelogram. 
For a square or parallelogram, there must be four points. Here, we only have three points.
So, they must form a triangle.