# 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 - Mathematics

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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

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#### Solution

Let A (–3, 0), B (1, –3) and C (4, 1) be the given points. Then,

AB=sqrt({1-(-3)}^2+(-3-0)^2)=sqrt(16+9)=

BC=sqrt((4-1)^2+(1+3)^2)=\sqrt{9+16}=

CA=sqrt((4+3)^2+(1-0)^2)=sqrt(49+1)=5\sqrt{2}

Clearly, AB = BC. Therefore, ∆ABC is isosceles.

Also,

AB^2 + BC^2 = 25 + 25 = (5)^2 = CA^2

⇒ ∆ABC is right-angled at B.

Thus, ∆ABC is a right-angled isosceles triangle.

Now,

Area of ∆ABC = \frac { 1 }{ 2 } (Base × Height)

= \frac { 1 }{ 2 } (AB × BC)

⇒ Area of ∆ABC = \frac { 1 }{ 2 } × 5 × 5 sq. units

= \frac { 25 }{ 2 } sq. units

Concept: Distance Formula
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