38. Find the area of a triangle ABC, whose vertices are A (-2,2), B (5,2) and 1 point
whose centroid is (1, 3).
1) 21/2
O2) 23/2
3) 19/2
4) 25/2
Answers
Step-by-step explanation:
Centroid of a triangle with vertices (x
1
,y
1
);(x
2
,y
2
) and (x
3
,y
3
) is calculated by the formula (
3
x
1
+x
2
+x
3
,
3
y
1
+y
2
+y
3
)
Let the third vertex of the triangle be C(x,y)
So, centroid =(
3
−2+5+x
,
3
2+2+y
)=(1,3)
=(
3
3+x
,
3
4+y
)=(1,3)
=>3+x=3;4+y=9
x=0;y=5
Hence the third vertex of the triangle is (0,5)
Area of a triangle with vertices (x
1
,y
1
) ; (x
2
,y
2
) and (x
3
,y
3
) is
∣
∣
∣
∣
2
x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)
∣
∣
∣
∣
Hence, substituting the points (x
1
,y
1
)=(−2,2) ; (x
2
,y
2
)=(5,2) and (x
3
,y
3
)=(0,5) in the area formula, we get
Area of triangle ABC =
∣
∣
∣
∣
2
(−2)(2−5)+(5)(5−2)+0(2−2)
∣
∣
∣
∣
=
∣
∣
∣
2
6+15
∣
∣
∣
=
2
21
squnits
I hope this will help you answer is 21/2
Answer:
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Step-by-step explanation:
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