Question

Triangle ABC has vertices at A(-2,1), B(-2,-3), and C(1,-2). What is the area of the triangle? a 18 square units b 9 square units c 6 square units d 12 square units

Answer 1

Answer:

Option C. 6 square units

Step-by-step explanation:

we know that

Heron's Formula is a method for calculating the area of a triangle when you know the lengths of all three sides.  

Let  

a,b,c be the lengths of the sides of a triangle.  

The area is given by:

$A=\sqrt{p(p-a)(p-b)(p-c)}$

where

p is half the perimeter

p=$\frac{a+b+c}{2}$

we have

Triangle ABC has vertices at A(-2,1), B(-2,-3), and C(1,-2)

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

$d=\sqrt{(-3-1)^{2}+(-2+2)^{2}}$

$d=\sqrt{(-4)^{2}+(0)^{2}}$

$dAB=4\ units$

step 2

Find the distance BC

$d=\sqrt{(-2+3)^{2}+(1+2)^{2}}$

$d=\sqrt{(1)^{2}+(3)^{2}}$

$dBC=\sqrt{10}\ units$

step 3

Find the distance AC

$d=\sqrt{(-2-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-3)^{2}+(3)^{2}}$

$dBC=\sqrt{18}\ units$

step 4

$a=AB=4\ units$

$b=BC=\sqrt{10}\ units$

$c=AC=\sqrt{18}\ units$

Find the half perimeter p

p=$\frac{4+\sqrt{10}+\sqrt{18}}{2}=5.70\ units$

Find the area

$A=\sqrt{5.7(5.7-4)(5.7-3.16)(5.7-4.24)}$

$A=\sqrt{5.7(1.7)(2.54)(1.46)}$

$A=\sqrt{35.93}$

$A=6\ units^{2}$