Raymond B. answered • 08/08/21
C is correct. Area with (-2,1) is about twice 6 = 12
it creates a near right triangle with Area =bh/2 = sqr5sqr29/2 = about 6.02, multiply it by 2 to get 12 for the parallelogram
even a rough sketch shows about 6 as the area, for the triangle if you use graph paper.
with (2,-1) the lines through the origin to the other vertices are y=2.5x and y=-(1/2)x, while not negative inverse slopes, they almost are, although it doesn't form a right 90 degree triangle, it's an 85.2 degree triangle, so the area is less than about 6.02
two sides of the triangle are sqr5 and sqr29, use the law of cosines to find the the 3rd side, c
when c^2 = a^2 +b^2 -2abCosC where C= 85.2.
c^2 = 5 + 29 -2sqr(5(29))Cos85.2
c = about 5.76
Now you have 3 sides of a triangle. Use Heron's formula to calculate area
Area = square root of [p(c-p)(b-p)(a-p)] where p is half the perimeter and a, b and c are the 3 side lengths.
It's about 6 again. Multiply by 2 to get the parallelogram's area = 12
do the same method to eliminate or confirm (-2,3), (-2,-7) and (-2, -11)
for (-2,3) the angle formed between the two lines is about 55.5 degrees
with sides sqr13 and sqr29.
3rd side square = 13+29-2(sqr13x29)Cos55.5
plot the points on a graph area is maybe 8 or so, times 2 is about 16
eliminate 3
that leaves y=-7 or -11
graph -7 and it just looks so narrow a triangle as to have nothing close to 12
the graph makes -11 unlikely too. It's just too narrow a triangle, with a nearly 180 degree angle.
third side would be about 16.5
Area = is too small