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