three vertices of a parallelogram ABCD are A(3,-4),B(-1,-3),C(-6,2) find the coordinates of vertex D and find the area of parallelogram ABCD
Answers
Answered by
4
Answer:
no need of this question
Answered by
5
Answer:Given A(3,−4),B(−1,−3),C(−6,2)
O is M.P of AC=(
2
3−6
,
2
−4+2
)=(−
2
3
,−1)
∴ O=(−
2
3
,−1)
Similarly, O is MP of BD
⇒
2
−3
=
2
−1+a
, −1=
2
−3+6
∴ a=−2,b=+1
∴ D=(−2,1)
∴ Area of parallelogram ABCD = Ar of △ABC + Ar △ADC
Area of △ABC=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
=
2
1
[3(−3−2)−1(2+4)−6(4+3)]
=
2
1
[−15−6+6]=
2
∣15∣
=
2
15
Area of △ADC=
2
1
[3(1−2)−2(2+4)−6(−4−1)]
=
2
1
[−3−12+30]=
2
15
∴ Area of parallelogram ABCD=
2
15
+
2
15
=15
Hope it helped u
plz mark me as brainliest
Thank u
Similar questions