Prove that the points (–3, 0), (1, –3) and (4, 1) are the vertices of an isosceles right angled triangle. Find the area of this triangle.
[4 MARKS]
Formula: 1 Mark
Concept of right-angled triangles: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Let A(-3, 0), B(1, -3) and C(4, 1) be the given points. Then,
Distance between the points is given by
√(x1−x2)2+(y1−y2)2
AB=√(1−(−3))2+(−3−0)2=√16+9=5 units.
BC=√(4−1)2+(1+3)2=√9+16=5 units.
AC=√(4+3)2+(1−0)2=√49+1=5√2 units.
Clearly, AB = BC. Therefore, ΔABC is isosceles.
Also,
AB2+BC2=25+25=(5√2)2=CA2
⇒ΔABC is right- angled at B because it is satisfying pythagoras theorem.
Thus, ΔABC is right- angled isosceles triangle.
Now, Area of ΔABC=12(Base×Height)
=12(AB×BC)
⇒Area of ΔABC=(12×AB×BC) sq. units
=252 sq. units.