Question

Find the area of the triangle formed by the origin and the points of intersection of the parabolas with equations y=−3x2+20 and y=x2−16 .

Answer 1

Answer:

area = 21 sq. units

Step-by-step explanation:

Answer 2

Answer:

Area = 21 unit²

Step-by-step explanation:

We have one point of triangle = (0,0)

The other points can be solved, intersection means

              −3x²+20 = x²−16

               4x² = 36

               x² = 9

               x = ±3  

when x = 3,      y =  3²−16  = 9 -16 = -7  , point is (3,-7)

when x = -3,      y =  (-3)²−16  = 9 -16 = -7  , point is (-3,-7)

We need to find area of triangle with points  (0,0), (3,-7) and (-3,-7).

Area of triangle is given by

         $Area =\frac{1}{2}\begin{vmatrix}0 & 0 & 1\\ 3 & -7 & 1\\ -3 & -7 & 1\end{vmatrix}$

  $Area = \frac{1}{2}\left ( 0(-7\times 1-(1\times -7))-0(3\times 1-(1\times -3))+1(3\times -7-(-3\times -7)\right )\\\\Area =\frac{1}{2}(-21-21)=-21=21unit^2$

Area = 21 unit²