Find the area of triangle whose vertices are 2, 3, 1, 0, 2 , -4
Answers
Answer:
Let A(2,3), B(−1,0) and C(2,−4) are the vertices of △ABC
Area of triangle=21[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
Here (x1,y1)=(2,3)
(x2,y2)=(−1,0)
(x3,y3)=(2,−4)
=21[2(0+4)−1(−4−3)+2(3−0)]
=21[8+7+6]
=221 sq.unit
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Answer:
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Class 11
>>Applied Mathematics
>>Straight lines
>>Introduction
>>Find the area of the triangle whose vert
Question
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Find the area of the triangle whose vertices are :
(i) (2,3),(−1,0),(2,−4)
(ii) (−5,−1),(3,−5),(5,2)
Medium
Solution
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(i) Let A(2,3), B(−1,0) and C(2,−4) are the vertices of △ABC
Area of triangle=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
Here (x
1
,y
1
)=(2,3)
(x
2
,y
2
)=(−1,0)
(x
3
,y
3
)=(2,−4)
=
2
1
[2(0+4)−1(−4−3)+2(3−0)]
=
2
1
[8+7+6]
=
2
21
sq.unit
(i) Let A(−5,1), B(3,−5) and C(5,2) are the vertices of △ABC
Area of triangle=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
Here (x
1
,y
1
)=(−5,−1)
=(x
2
,y
2
)=(3,−5)
=(x
3
,y
3
)=(5,2)
=
2
1
[−5(−5−2)+3(2+1)+5(−1+5)]
=
2
1
[35+9+20]
=
2
64
=32 sq.unit
Step-by-step explanation:
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