What is the area of a triangle whose vertices are the points with coordinates (3,2) (5,10) and (8,4)?

1st solution

We can use Heron formula which states

The area of a triangle with sides a,b,c is equal to

#S=sqrt(s(s-a)(s-b)(s-c))# where #s=(a+b+c)/2#

No using the formula to find the distance between two points
#A(x_A,y_A) , B(x_B,y_B)#which is

#(AB)=sqrt((x_A-x_B)^2+(y_A-y_B)^2#

we can calculate the length of sides between the three points given
let say #A(3,2)# #B(5,10)# , #C(8,4)#

After that, we substitute to Heron formula.

2nd Solution

We know that if #(x_1,y_1), (x_2,y_2)# and #(x_3,y_3)# are the vertices of the triangle, then the area of the triangle is given by:

Area of the triangle# = (1/2) |{(x2-x1)(y2+y1) +(x3-x2)(y3+y1)+(x1-x3)(y1+y2)}|#

Therefore the area of the triangle whose vertices are #(3,2), (5,10), (8 ,4)# is given by:

Area of the triangle# = (1/2) |{(5-3)(10+2) +(8-5)(4+2)+(3-8)(2+10)}|=abs(1/2(24+18-60))=9#

Method 1: Geometric

#triangle ABC = PQRS - (triangleAPB+triangleBQC+ACRS)#

#PQRS = 5xx10 = 50#
#triangle APB = 1/2(8xx2) = 8#
#triangle BQC = 1/2(3xx6) =9#
#ACRS = (2+4)/2xx5 =15#

#triangle ABC = 50 -(8+9+15) = 50 -32 = 18#

Method 2: Herons Formula
Using the Pythagorean Theorem we can calculate the lengths of the sides of #triangle ABC#
then we can use Heron's Formula for the area of a triangle given the lengths of its sides.

Because of the number of calculations involved (and the need to evaluate square roots), I did this in a spreadsheet:

Again (fortunately) I got an answer of #18# for the area