How can I find the area of a triangle with these three given points?

Question

The points are (-1,0) (2,7) and (2,0)

Answer 1

Step 1) Find the sides length of triangle.

Given points are A(-1,0),B(2,7), C(2,0)

Now find the each distance of between two points with the formula √[(x2-x1)^2+(y2-y1)^2]

AB = √[(2+1)^2+(7-0)^2]

= √(9+49) = √58

BC = √[(2-2)^2+(0-7)^2]

= √(0+49) = 7

CA = √[(-1-2)^2+(0-0)^2]

= √9+0 = 3

Distances are denote with a,b,c.

Step 2) Area of triangle A = √[S(S-a)(S-b)(S-c)]

Here S is semiperimeter.

S = (a+b+c)/2

S = (√58+7+3)/2

S = (√58+10)/2

S = √58/2+5

A = √[√58/2+5(√58/2+5-√58)(√58/2+5-7)(√58/2+5-3)]

A = √[√58/2+5((√58+10-2√58)/2)((√58-4)/2)((√58+4))/2]

A =√ [√58/2+5(10-√58/2)((√58)^2-(4)^2/4]

A = √[√58/2+5(10-√58/2)(58-16)/4]

A = √[(10+√58)/2(10-√58)/2(42/4)]

A = √[ (10^2-(√58)^2)/4(42/4)]

A = √[(100-58)42/16]

A = √(42*42/4*4)

A = 42/4

A = 21/2

A = 10.5