Question

show that the point A (0,9),B(-4,-1),C(3,2)are the vertices of a right angled triangle
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Answer 1

S O L U T I O N :

Given :- Vertices of a right angled triangle is A(0,9),B(-4,-1),C(3, 2).

Applying distance formula,

XY = √(x2 - x1)² + (y2 - y1)²

Case (I),

A(0,9) & B(-4 ,-1)

⟶ AB = √(-4 - 0)² + (-1 - 9)²

⟶ AB = √(-4)² + (-10)²

⟶ AB = √16 + 100

⟶ AB = √116___________(1)

Case (II),

B(-4, -1) & C(3 , 2)

⟶ BC = √(3 - (-4))² + (2 - (-1))²

⟶ BC = √(3 + 4)² + (2 + 1)²

⟶ BC = √(7)² + (3)²

⟶ BC = √49 + 9

⟶ BC = √58___________(2)

Case (III),

A(0,9) & C(3 , 2)

⟶ AC = √(3 - 0)² + (2 - 9)²

⟶ AC = √(3)² + (-7)²

⟶ AC = √9 + 49

⟶ AC = √58_____________(3)

From equation (1),

⟶ AB² = (√116)² = 116

⟶ AB² = 116_____________(4)

From equation (2) & (3),

⟶ AC² + BC² = (√58)² + (√58)²

⟶ AC² + BC² = 58 + 58

⟶ AC² + BC² = 116___________(5)

From equation (4) & (5),

⟶ AB² = AC² + BC²

By using converse of Pythagoras theorem,

⟶ ΔACB is right angled triangle.

Therefore,

• The point A(0,9) , B(-4,-1) , C(3,2) are the vertices of a right angled triangle.