Question

find the area of the triangle whose vertices are (2,3),(-1,0),(2,-4)

full step needed

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

Answer 2

Answer:

$\frac{21}{2}\text{square units}$ is the required area of triangle.

Step-by-step explanation:

We have been given the vertices of triangle

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

$Area=\frac{1}{2}[x_1(y2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$

Here,$x_1=2,y_1=3,x_2=-1,y_2=0,x_3=2,y3=-4$

On substituting the values in the formula we get:

$\frac{1}{2}[2(0+4)-1(-4-3)+2(3-0)]$

$\frac{1}{2}[8+7+6]$

$\frac{21}{2}\text{square units}$.