What is the area of triangle ABC with vertices A(2, 3), B(1, -3), and C(-3, 1)?

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What is the area of triangle ABC with vertices A(2, 3), B(1, -3), and C(-3, 1)?

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Answer 1 · 2016-11-04T04:10:08

Area = 14 square units

Explanation:

First, after applying the distance formula $a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$ , we find that side length opposite to point A (call it $a$ $a$ ) $a = 4 \sqrt{2}$ $a=4sqrt2$ , $b = \sqrt{29}$ $b=sqrt29$ , and $c = \sqrt{37}$ $c=sqrt37$ .
Next, use Herons rule:
$A r e a = \sqrt{s (s - a) (s - b) (s - c)}$ $Area = sqrt(s(s-a)(s-b)(s-c))$ where $s = \frac{a + b + c}{2}$ $s=(a+b+c)/2$ .
We then get:
$A r e a = \sqrt{(2 \sqrt{2} + \frac{1}{2} \sqrt{29} + \frac{1}{2} \sqrt{37}) (- 2 \sqrt{2} + \frac{1}{2} \sqrt{29} + \frac{1}{2} \sqrt{37}) (2 \sqrt{2} - \frac{1}{2} \sqrt{29} + \frac{1}{2} \sqrt{37}) (2 \sqrt{2} + \frac{1}{2} \sqrt{29} - \frac{1}{2} \sqrt{37})}$ $Area = sqrt[(2sqrt2+1/2sqrt29+1/2sqrt37)(-2sqrt2+1/2sqrt29+1/2sqrt37)(2sqrt2-1/2sqrt29+1/2sqrt37)(2sqrt2+1/2sqrt29-1/2sqrt37)]$
It's not as scary as it looks. This simplifies down to:
$A r e a = \sqrt{196}$ $Area = sqrt196$ , so $A r e a = 14$ $Area =14$ $u n i t s^{2}$ $units^2$

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What is the area of triangle ABC with vertices A(2, 3), B(1, -3), and C(-3, 1)?

1 Answer

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