Question

prove that the points (-3,0),(1,-3) and (4,1) are the vertices of an isosceles right-angled triangle.

Answer 1

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.

Answer 2

Distance formula = √(x2 -x1)^2 + (y2-y1)^2

A = (-3,0) ; B = (1,-3) ; C = (4,1)

AB = √( 1-(-3) )^2 +( -3-0 )^2
= √16 + 9
= √25
= 5

BC = √(4-1)^2 + (1+3)^2
=√9 + 16
= √25
= 5

CA = √(4+3)^2+(1-0)^2
=√14+1
=√15

since, AB=BC

THEREFORE ,
It is an isosceles triangle as the two sides are same.(proved)

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