Question

Prove that the triangle ABC with vertices A (0,4), B (-1,0) and C (3,-1) is an
isosceles right angled triangle.

Answer 1

Answer:

We know that the distance between the two points (x 1 ,y 1 ) and (x 2 ,y 2 ) is  

d= (x 2 −x 1 ) 2 +(y 2 −y 1 ) 2

Let the given vertices be A=(1,−3), B=(−3,0) and C=(4,1)

We first find the distance between A=(1,−3) and B=(−3,0) as follows:

AB= (x 2 −x 1 ) 2 +(y 2 −y 1 ) 2

= (−3−1) 2 +(0−(−3)) 2

= (−4) 2 +(0+3) 2

= (−4) 2 +3 2

= 16+9 = 25 =5

Similarly, the distance between B=(−3,0) and C=(4,1) is:

BC= (x 2 −x 1 ) 2 +(y 2−y 1 ) 2 = (4−(−3)) 2 +(1−0) 2  

= (4+3) 2 +1 2  

= 7 2 +1 2 = 49+1 = 50 = 5 2 ×2 =5 2

Now, the distance between C=(4,1) and A=(1,−3) is:

CA= (x 2 −x 1 ) 2+(y 2 −y 1 ) 2 = (1−4) 2 +(−3−1) 2 = (−3) 2 +(−4) 2 = 9+16 = 25 = 5 2 =5

We also know that If any two sides have equal side lengths, then the triangle is isosceles.

Here, since the lengths of the two sides are equal that is AB=CA=5, therefore, ABC is an isosceles triangle.

Now consider,

AB 2 +CA 2  

=5 2 +5 2 =25+25=50=(5 2 ) 2 =BC 2

Since AB 2 +CA 2=BC 2

, therefore, ABC is a right angled triangle.

Hence, the given vertices are the vertices of isosceles right angled triangle.

Step-by-step explanation:

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