Question

Find the area of the triangle formed by the three points whose coordinates are (2, 3), (4, 5) and (6, 3).

Answer 1

Answer:

Step-by-step explanation:Firstly let's find the length of each AB,BC,AC

A(2;3) B(4;5) C(6;3)

AB=√ (xA-xb)²+ (yA-yB)²= √(2-4)² + (3-5)²= 2√2

BC= √(xB-xC)²+(yB-yC)²=√(4-6)²+(5-3)²= 2√2

AC=√(xA-xC)²+(yA-yC)²=√ (2-6)²+(3-3)²= 4

Now let s have a look

AC²= 4²= 16

AB²+ BC²= (2√2)²+(2√2)²= 8+8=16

As AC²=AB²+BC² the ABC triangle is right angled

The area - A= (BC*AB)/2= (2√2+2√2)/2=4√2/2=2√2