A(2,1),B(8,1)andC(8,9) perimeter of a triangle
Answers
Answer:
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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=(−2,1), B=(4,6) and C=(6,−3)
We first find the distance between A=(−2,1) and B=(4,6) as follows:
AB=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(4−(−2))
2
+(6−1)
2
=
(4+2)
2
+5
2
=
6
2
+5
2
=
36+25
=
61
Similarly, the distance between B=(4,6) and C=(6,−3) is:
BC=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(6−4)
2
+(−3−6)
2
=
2
2
+(−9)
2
=
4+81
=
85
Now, the distance between C=(6,−3) and A=(−2,1) is:
CA=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(6−(−2))
2
+(−3−1)
2
=
(6+2)
2
+(−4)
2
=
8
2
+(−4)
2
=
64+16
=
80
Since the perimeter P of a triangle ABC is AB+BC+CA, therefore,
P=
61
+
85
+
80
Hence, the perimeter of the triangle is (
61
+
85
+
80
) units.