Question Number 100543 by mhmd last updated on 27/Jun/20

Commented bymr W last updated on 27/Jun/20

(1) y^2 =4+4−2y ⇒y^2 +2y−8=0 Δ=2^2 +4×8=36 Area=((Δ(√Δ))/(6a^2 ))=((36(√(36)))/(6×1^2 ))=36 (4) 2y^2 +2x=0 ⇒2y^2 +y−1=0 Δ=1^2 +4×1×1=5 Area=((Δ(√Δ))/(6a^2 ))=((5(√5))/(6×2^2 ))=((5(√5))/(24)) see Q89781

Answered by Rio Michael last updated on 27/Jun/20

1 ↘↙ 6 12 2 ↘↙ 4 = 4 −12 −3 ↘↙ 7 = 14 7 1 ↘ ↙ 6 = −18 ⇒ Area of triangle = (1/2)[4 + 14−18 −(12−12 + 7)] = (7/2) square units

Answered by Rio Michael last updated on 27/Jun/20

Curves intersect at the points (−2,5) , (2.3,0.7) and (1.25,−1.50) ⇒ −2 5 16 2.3 0.7 −1.4 0.851.25 −1.5 3.45 3 −2 5 6.25 A = (1/2)[−1.4+3.45 + 6.25 −(16 + 0.85 + 3)] ≈ 5.78 square units