The area of the triangle is 10 sq units
SOLUTION:
Given, we have to find the area of a triangle whose vertices are D(3, 3), E(3, −1) and F(−2, −5)
We know that,
[tex]\text { Area of triangle }=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right][/tex]
Where, [tex]\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right),\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)[/tex] are vertices of the triangle.
Here in our problem, [tex]\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(3,3),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(3,-1) \text { and }\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)=(-2,-5)[/tex]
Now, substitute the above values in the formula.
[tex]\begin{aligned} \text { Area of triangle } &=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right| \\\\ &=\frac{1}{2}|3(-1-(-5))+3(-5-3)+(-2)(3-(-1))| \\\\ &=\frac{1}{2}|3(-1+5)+3(-8)-2(3+1)| \\\\ &=\frac{1}{2}|3(4)-24-2(4)| \\\\ &=\frac{1}{2}|12-24-8| \\\\ &=\frac{1}{2}|12-32|=\frac{1}{2}|-20|=\frac{20}{2}=10 \text { sq units } \end{aligned}[/tex]
