Question

Prove that the points 3, 0, 6, 4 and 13 are the vertices of a right angled isosceles triangle

Answer 1

Answer:

Explained below

Step-by-step explanation:

Let A, B and C be the point of Given Triangle, where A (3,0),  B (6,4) and C (1,3)

Now using distance formula,

AB=√ {(6-3)²+(4-0)²}= √(9+16)=√25= 5

 BC=√{(1-6)²+(3-4)²}=√(25+1)=√26

CA =√{(1-3)²+(3-0)²}=√(4+9)=√13

This is not a isosceles triangle.

However, If point C contains the values C(-1,3) then

BC=√{(-1-6)²+(3-4)²}=√(49+1)=√50

CA =√{(-1-3)²+(3-0)²}=√(16+9)=√25=5

Here, AB=AC

Now using Pythagoras theorem

hypotenuse²= √sum of squares of other two sides.

BC² =AC² + AB²

(√50)²=√(5²+5²)=50

50=50

which is related to Pythagoras theorem .

Therefore, ABC are the vertices of right angle isosceles triangle with ∠A at right angle.