Question

Find the area of triangle jkl above if the vertices are j(2,2), k(3,5), and L(6,2)

Answer 1

Answer:

$Area=6\:units^2$

Step-by-step explanation:

Given the vertices $(x_1,y_1)=J(2,2)$,  $(x_2,y_2)=K(3,5)$ and

$(x_3,y_3)=L(6,2)$. The area of triangle JKL is given by:

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

We substitute the values to get:

$Area=\frac{1}{2}|2(5-2)+3(2-2)+6(2-5)|$

This simplifies to

$Area=\frac{1}{2}|2(3)+3(0)+6(-3)|$

$Area=\frac{1}{2}|6+0-18|$

$Area=\frac{1}{2}|-12|$

$Area=\frac{1}{2}*12$

$Area=6\:units^2$