Area given vertices

Geometry Level 2

Find the area of a triangle whose vertices have coordinates (2,3), (-4,2), (10,1).


The answer is 10.

View wiki
April Queen Piedad by April Queen Piedad , 29, Philippines

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

A = 1 2 ∣ x 1 x 2 . . . x n x 1 y 1 y 2 . . . y n y 1 ∣ = 1 2 ∣ 2 − 4 10 2 3 2 1 3 ∣ = 1 2 [ 4 − 4 + 30 − ( − 12 + 20 + 2 ) ] = 1 2 ( 20 ) = A=\dfrac{1}{2} \begin{vmatrix} x_1 & x_2 ... & x_n & x_1 \\ y_1 & y_2 ... & y_n & y_1 \end{vmatrix} =\dfrac{1}{2}\begin{vmatrix} 2 & -4 & 10 & 2 \\ 3 & 2 & 1 & 3 \end{vmatrix}=\dfrac{1}{2}[4-4+30-(-12+20+2)]=\dfrac{1}{2}(20)= 10 \boxed{10}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Jonathan Yang
Feb 25, 2015

By Shoelace theorem, the area is ( 2 ∗ 2 − 4 ∗ 1 + 10 ∗ 3 ) − ( − 4 ∗ 3 + 10 ∗ 2 + 2 ∗ 1 ) 2 = 10 \frac{(2*2-4*1+10*3)-(-4*3+10*2+2*1)}{2} = 10

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...