If 3a+5abc-3c=6d find ‘a’ when b=5,c= -2, and d=4 please full explanation! and please fast!!! About the author Peyton
Answer: Given, A(6,3),B(−3,5),C(4,−2)andD(x,3x) By using area of triangle = 2 1 ∣x 1 (y 2 −y 3 )+x 2 (y 3 −y 2 )+x 3 (y 1 −y 2 )∣ Area of ΔDBC = ∣ ∣ ∣ ∣ ∣ 2 (x)(5+2)+(−3)(−2−3x)+4(3x−5) ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ 2 7x+6+9x+12x−20 ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ 2 28x−14 ∣ ∣ ∣ ∣ ∣ Area of ΔABC = ∣ ∣ ∣ ∣ ∣ 2 (6)(5+2)+(−3)(−2−3)+4(3−5) ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ 2 42+15−8 ∣ ∣ ∣ ∣ ∣ = 2 49 Given: AreaofABC AreaofDBC = 2 1 2 49 ∣ ∣ ∣ ∣ ∣ 2 28x−14 ∣ ∣ ∣ ∣ ∣ = 2 1 ∣28×14∣= 2 49 28×14=± 2 49 Taking a positive sign 28×14= 2 49 56x−28=49 56x=77 x= 56 77 = 8 11 Taking a negative sign 28×14=− 2 49 56x−28=−49 56x=−21 x= 56 21 = 8 3 So, x= 8 11 or x= 8 3 Reply
Answer:
Given,
A(6,3),B(−3,5),C(4,−2)andD(x,3x)
By using area of triangle =
2
1
∣x
1
(y
2
−y
3
)+x
2
(y
3
−y
2
)+x
3
(y
1
−y
2
)∣
Area of ΔDBC =
∣
∣
∣
∣
∣
2
(x)(5+2)+(−3)(−2−3x)+4(3x−5)
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
2
7x+6+9x+12x−20
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
2
28x−14
∣
∣
∣
∣
∣
Area of ΔABC =
∣
∣
∣
∣
∣
2
(6)(5+2)+(−3)(−2−3)+4(3−5)
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
2
42+15−8
∣
∣
∣
∣
∣
=
2
49
Given:
AreaofABC
AreaofDBC
=
2
1
2
49
∣
∣
∣
∣
∣
2
28x−14
∣
∣
∣
∣
∣
=
2
1
∣28×14∣=
2
49
28×14=±
2
49
Taking a positive sign
28×14=
2
49
56x−28=49
56x=77
x=
56
77
=
8
11
Taking a negative sign
28×14=−
2
49
56x−28=−49
56x=−21
x=
56
21
=
8
3
So, x=
8
11
or x=
8
3