c5 = x + 2 y = 1 = − 3 x = − 2 y = 5 =5 = x + 2 y = 1 = − 3 x = − 2 y = 5 =5 = x + 2 y = 1 = − 3 x = − 2 y = 5 =
Answers
Answered by
8
Answer:
Using ar(Δ)=
2
1
∣x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)∣
ar(ΔABD)=
2
1
∣(−3)(−2+4)+(−1)(−4−4)+x(4−(−2))∣
=
2
1
∣2+6x∣
=∣1+3x∣⟶(i)
ar(ΔACD)=
2
1
∣(−3)(6+4)+5(−4−4)+x(4−6)∣
=
2
1
∣−70−2x∣
=
2
1
∣70+2x∣
=35+x⟶(ii)
∵∣−1∣=1
So,
ar(ΔABD)=2Δ(ACD)
1+3x=70+2x
69=x
∴x=69
Answered by
1
Answer:
Using ar(Δ)=
2
1
∣x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)∣
ar(ΔABD)=
2
1
∣(−3)(−2+4)+(−1)(−4−4)+x(4−(−2))∣
=
2
1
∣2+6x∣
=∣1+3x∣⟶(i)
ar(ΔACD)=
2
1
∣(−3)(6+4)+5(−4−4)+x(4−6)∣
=
2
1
∣−70−2x∣
=
2
1
∣70+2x∣
=35+x⟶(ii)
∵∣−1∣=1
So,
ar(ΔABD)=2Δ(ACD)
1+3x=70+2x
69=x
∴x=69
Explanation:
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