Question

(1, 0), (0, 1), (1, 1) are the coordinates of vertices of a triangle. The triangle is ..... triangle.select a proper option (a), (b), (c) or (d) from given options so that the statement becomes correct.
(a) acute angled
(b) obtuse angled
(c) isosceles
(d) equilateral

Answer 1

Let A = (1,0) , B = (0,1) and C = (1, 1)
use distance formula to find side length of AB, BC and CD of triangle ABC.

AB = $\sqrt{(1-0)^2+(0-1)^2}=\sqrt{2}$
BC = $\sqrt{(1-0)^2+(1-1)^2}=1$
CA = $\sqrt{(1-1)^2+(0-1)^2}=1$

we can see that AB² = 2
BC² + CA² = 1 + 1 = 2
so, AB² = BC² + CA²
from Pythagoras theorem, ∆ABC is right amgled triangle where C is right angle .

but option is not given.
here we see one more clue .
BC = CA = 1
if two sides are equal then triangle is isosceles triangle.

hence, option (c) is correct.

Answer 2

Let A( 1 , 0 ) , B( 0, 1 ) and C( 1 , 1 ) are

the coordinates of veriticies of a

Triangle ABC.

By using midpoint formula :

i ) A( 1 , 0 ) = ( x1 , y1 )

B( 0 , 1 ) = ( x2 , y2 )


AB = √ ( x2 - x1 )² + ( y2 - y1 )²

= √(0-1)² + (1-0)²

= √ 1 + 1

AB = √2 ----( 1 )


ii ) B( 0,1 ) , C( 1 ,1 )

BC = √ ( 1-0)² + ( 1 - 1 )²

= √ 1 + 0

BC = 1 -----( 2 )

iii ) C( 1 , 1 ) , A( 1 , 0 )

CA = √( 1 - 1 )² + ( 0 - 1 )²

= √ 0 + 1

= 1

CA = 1 -----( 3 )

From ( 1 ) , ( 2 ) and ( 3 ),

iv ) BC = CA = 1

v ) AB² = BC² + CA²

=> ( √2 )² = 1² + 1²

=> 2 = 2 ( True )

Therefore ,

Triangle ABC is a Isosceles right

angled triangle .

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