Question

find the area of a triangle ABC whose vertices are a (2,3 )b (- 1, 0 )and c (2, - 4)​

Answer 1

A(2,3) B(-1,0) C(2,-4)

X1=2         X2=-1        X3=2

Y1=3          Y2=0        Y3= -4

1/2  {x1(y2-y3) + x2(y3-y1) + x3(y1-y2)}

1/2 {2(0-(-4)) + (-1) (-4-3) + 2(3-0)}

1/2 {8-(-7)+6}

1/2 {8+7+6}

1/2 * 21

21/2

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Answer 2

Answer:

Area  =  21 / 2  =  10.5

Step-by-step explanation:

Method 1  --  Using the 3×3 determinant method

$\displaystyle\text{Area}=\tfrac12\left|\ \left|\begin{array}{ccc}1&1&1\\2&-1&2\\3&0&-4\end{array}\right|\ \right|\\\\{}\qquad=\tfrac12\left|\ \left|\begin{array}{cc}-1&2\\0&-4\end{array}\right|-\left|\begin{array}{cc}2&2\\3&-4\end{array}\right|+\left|\begin{array}{cc}2&-1\\3&0\end{array}\right|\ \right|\\\\{}\qquad=\tfrac12\left|4 +14+3\right|\\\\{}\qquad=\tfrac{21}{2}=10.5$

Method 2  --  Being observant

The points a and c have the same x-coordinate, so take the side ac as "base".  The length of the base is then the difference between the y-coordinates of a and c, so:

base = 3 - -4 = 3 + 4 = 7

The "height" is then the distance that b=(-1,0) is from the base line x=2, so:

height = 2 - -1 = 2+1 = 3

So...

Area = (1/2) × base × height

       = (1/2) × 7 × 3

       = 21 / 2