Question

prove that the points (10,10),(10,0) and (0,0) are the vertices of an isosceles triangle​

Answer 1

Given :-

• The given points are ( 10 , 10 ) , ( 10 , 0 ) and ( 0,0 )

Solution :-

Let consider the points A, B and C

The coordinates of point A = ( 10 , 10 )

The coordinates of point B = ( 10 , 0 )

The coordinates of point C = ( 0 , 0 )

Let consider the triangle ABC

Therefore,

The sides of triangle are AB, BC and AC

By using Distance formula ,

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

AB = √( 10 - 10 )^2 + ( 0 - 10 )^2

AB = √ 0 + 100

AB = √ 100

AB = 10

Now,

BC = √ ( 0 - 10 )^2 + ( 0 - 0 )^2

BC = √100 + 0

BC = √100

BC = 10

Now,

AC = √ ( 0 - 10)^2 + ( 0 - 10)^2

AC = √ 100 + 100

AC = √200

AC = 10√2

Here, Side AB = Side BC

Hence, It is isosceles triangle $.$

Answer 2

Answer:

Answer. So length of each equal side is 11cm.

Step-by-step explanation:

So it's a isosceles triangle