Question

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

Answer 1

Answer:

By Distance Formula.

Step-by-step explanation:

let the vertices of triangle be A,B,C.

$\sqrt{(x1 - x2) {}^{2} }+ \sqrt{ ( y2 - y1) {}^{2} }$

$\sqrt{(3 - 6) {}^{2} + (0 - 4) {}^{2} }$

$\sqrt{9 + 16}$

$\sqrt{25}$

$5$

AB=5

Answer 2

Step-by-step explanation:

Question:

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

Solution:

Given-

A(3, 0), B(6, 4) and C(-1, 3)

=> AB² = (3-6)²+ (0-4)²

=> AB² = 9+ 16

=> AB² = 25

=> BC² = (6 + 1)² + (4 - 3)²

=> BC² = 49 + 1

=> BC² = 50

Now,

= > CA² = (-1 - 3)² + (3-0)²

=> CA² = 16 + 9

=> CA² = 25

Hence,

AB² = CA² => AB = CA

It means triangle is isosceles

Also,

=> 25 + 25 = 50

=> AB² + CA² = BC²

Since Pythagoras theorem is verified, therefore triangle is a right angled triangle.