Question
Find the area of a triangle whose vertices are,
$(6, 3), (-3, 5)$ and $(4, -2).$
$(6, 3), (-3, 5)$ and $(4, -2).$
✓
Answer
Area of a triangle is given by,$\frac{1}{2}\big[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big]$
Here, $x_1 = 6, y_1 = 3, x_2 = -3, y_2 = 5, x_3 = 4, y_3 = -2$
Let A(6, 3), B(-3, 5) and C(4, -2) be the given points
Area of $\triangle\text{ABC}=\frac{1}{2}\Big[6(5+2)+(-3)(-2-3)+4(3-5)\Big]$
$=\frac{1}{2}\Big[6\times7-3\times(-5)+4(-2)\Big]$
$=\frac{1}{2}\Big[42+15-8\Big]$
$=\frac{49}{2}\text{sq.units}$
Here, $x_1 = 6, y_1 = 3, x_2 = -3, y_2 = 5, x_3 = 4, y_3 = -2$
Let A(6, 3), B(-3, 5) and C(4, -2) be the given points
Area of $\triangle\text{ABC}=\frac{1}{2}\Big[6(5+2)+(-3)(-2-3)+4(3-5)\Big]$
$=\frac{1}{2}\Big[6\times7-3\times(-5)+4(-2)\Big]$
$=\frac{1}{2}\Big[42+15-8\Big]$
$=\frac{49}{2}\text{sq.units}$
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