Question

Prove that points (3,0) , (6,4) and (-1,3) are vertices of a right angled isosceles triangle.

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Answer 1

rules of an isoceles triangle:

two sides are equal

Given,

A= 3,0

B= 6,4

C= -1,3

distance formula = √ (X2 - X1)^2 + (Y2- Y1)^2

Therefore

AB = √ (6-3)^2 + 4^2

= √ 9+ 16

= √25

= 5

BC = √(-1-6)^2 +(3-4)^2

= √ 49+1

= 5√2

CA = √(3+1)^2 +3^2

= √ 16 +9

= 5

therefore. : AB = CA..

since two sides are equal in length. It is an isosceles triangle

Answer 2

Answer:

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Step-by-step explanation:

AB and BC are the isosceles sides.

By distance formula, AB=(−1−3)2+(3−0)2=16+9=5

                                   BC=(3−6)2+(0−4)2=9+16=5

Since AB=BC, the ΔABC is isosceles.

For proving ∠ABC is right, we use the relation m1×m2=−1

where m1 is slope of line AB and m2 of BC :

m1=−1−43−0=−43   m2=6−34−0=34

m1×m2=4−3×34=−1