prove that A(2,3) B(-1,-1) C(3,-4) is isosceles triangle
Answers
Step-by-step explanation:
We know that the distance between the two points (x
1
,y
1
) and (x
2
,y
2
) is
d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Let the given vertices be A=(1,−3), B=(−3,0) and C=(4,1)
We first find the distance between A=(1,−3) and B=(−3,0) as follows:
AB=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(−3−1)
2
+(0−(−3))
2
=
(−4)
2
+(0+3)
2
=
(−4)
2
+3
2
=
16+9
=
25
=5
Similarly, the distance between B=(−3,0) and C=(4,1) is:
BC=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(4−(−3))
2
+(1−0)
2
=
(4+3)
2
+1
2
=
7
2
+1
2
=
49+1
=
50
=
5
2
×2
=5
2
Now, the distance between C=(4,1) and A=(1,−3) is:
CA=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(1−4)
2
+(−3−1)
2
=
(−3)
2
+(−4)
2
=
9+16
=
25
=
5
2
=5
We also know that If any two sides have equal side lengths, then the triangle is isosceles.
Here, since the lengths of the two sides are equal that is AB=CA=5, therefore, ABC is an isosceles triangle.
Now consider,
AB
2
+CA
2
=5
2
+5
2
=25+25=50=(5
2
)
2
=BC
2
Since AB
2
+CA
2
=BC
2
, therefore, ABC is a right angled triangle.
Hence, the given vertices are the vertices of isosceles right angled triangle.