In a 2D coordinate system point A(3,-6) B(1,1) & C(7,6). Find the area of the triangle ABC.?

Question

A. 26
B. 24
C. 20
D. 17.5
E. 14

Answer 1

A = 0.5*(x1*y2 + x2*y3 + x3*y1 - y1*x2 - y2*x3 - y3*x1)

Plug in your #s and find the result

I don't do arithmetic here

Answer 2

Herons formula works, but involves taking the square root, which is infact unecessary.

The easiest way to find the area without calculator is to find the area of the bounding box and then subtract the area of the (potentially 3) triangles surrounding the triangle in that bounding box. The "bounding box" the the box that has at the top right most coordinate the maximum x, y values and at the lower left the minimum(x,y) coordinate.

Area of bounding box is x_l = 7-1 = 6 and y_l = 6 - (-6) = 12
6*12 = 72

The other three triangles are determined by comparing to the maximum and minimum coordinates
(7,6) (7,-6), (1,6), (1,-6).

its easiest (for me) just to plot out a quick graph and figure out the traiangles involved, but the coordinates should be (In no particular order).

X:(7,6), (7,-6), (3,-6)
Y:(3,-6),(1,-6), (1,1)
Z:(1,1), (1,6), (7,6)

(Note how I proceeded ina clockwise fashion to get all triangles.
If a whole side of the triangle abc is collinear with one of the sides of the bounding box you'd have 1 less triangle).

the nice thing about these triangles is that the width and height are easy to calculate.

since the formula for a triangle is w*h/2 when you know width and height, we'lll just combine all the widths and heights together
and then divide by

X: w = 4 h = 12 w*h= 48
Y: w =2 h= 7 w*h = 14
Z w = 6 h = 5 w*h = 30


(48 + 14 + 30 )/2 = 46 = area of 3 remaning triangles.

which leaves the answer as ....?

Answer 3

use a determinant method i.e.
3 -6 1
1 1 1
7 6 1
solve it n the ans is 52
and the area is= 1/2 * (absolute value of 52)
which is equal to 26
hence area is A. 26

Answer 4

area = 26 units

Answer 5

Find the lengths of AB, AC, and BC and then plug into Hero's or Herron's formula (the same formula).

http://mathworld.wolfram.com/HeronsFormula.html