Prove that the point A (-3,0),B (1,-3),C(4,1) are the vertices of an isosceles right-angled triangle.
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Sol:
Vertices of the triangle are A(-3, 0), B(1, -3), C(4, 1)
Distance between two points = (x2 - x1)2+(y2 - y1)2
AB = √[(1 + 3)2 + (-3 -0)2] = 5
BC = √[(4 - 1)2 + (1 + 3)2] = 5
AC = √[(4 + 3)2 + (1 -0)2] = 50 = 52
AB = BC
Therefore, ΔABC is an isosceles triangle.
52 + 52 = (52 )2
⇒ 50 = 50
⇒ (AB)2 + (BC)2 = (AC)2
So, the triangle satisfies the pythagoras theorem and hence it is a right angled triangle.
Area of the isosceles right angled triangle = 1 2 ×base × height = 1 2 × 5 × 5 = 12.5 sq units.
R
Vertices of the triangle are A(-3, 0), B(1, -3), C(4, 1)
Distance between two points = (x2 - x1)2+(y2 - y1)2
AB = √[(1 + 3)2 + (-3 -0)2] = 5
BC = √[(4 - 1)2 + (1 + 3)2] = 5
AC = √[(4 + 3)2 + (1 -0)2] = 50 = 52
AB = BC
Therefore, ΔABC is an isosceles triangle.
52 + 52 = (52 )2
⇒ 50 = 50
⇒ (AB)2 + (BC)2 = (AC)2
So, the triangle satisfies the pythagoras theorem and hence it is a right angled triangle.
Area of the isosceles right angled triangle = 1 2 ×base × height = 1 2 × 5 × 5 = 12.5 sq units.
R
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