Find area of triangle whose vertices are (1,2),(-6,1),(0,8)
Answers
Answered by
2
Step-by-step explanation:
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
Similar questions