Question

show that the point A(8,3),B(0,9) and C(14,11) are the vertices of an isosceles right angled triangle

Answer 1

first find AB
AB=√(0-8)^2 +(9-3)^2 ,
=√(-8)^2 +(6)^2
=√64+36
=√100
AB=10.

BC=√(14-0)^2 +(11-9)^2
=√(14)^2 + (2)^2
=√196+4
=√200

CA=√(8-14)^2 + (3-11)^2
= √(-6)^2 + (-8)^2
=√36+64
=√100
=10

two sides of a triangle are equally so they are the vertices of an isosceles triangle.

comment me was this answer help u in solving ur problem.

Answer 2

Answer:

Yeah Both are equal

Step-by-step explanation:

(8,3) (0,9)

$x_{1} ,y_{1} x_{2} ,y_{2}$

$\sqrt{(x_{2}- y_{1})^{2} -(y_{2} -y_{1})^2 }$

$\sqrt{(0-8)^{2} -(9 -3)^2 }$

$\sqrt{100}$

10

you should do this and you will get the other 3 sides as, $10\sqrt{2}$, $10\sqrt{2}$, 10 and since two sides are equal it is an isoceles triangle