The area of a triangle having vertices A(3, 0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(B) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Given vertices are $A(3, 0)$,$B(0, 3)$,and $C(3, 3)$.Substituting these values into the formula:Area $= \frac{1}{2} |3(3 - 3) + 0(3 - 0) + 3(0 - 3)|$Area $= \frac{1}{2} |3(0) + 0(3) + 3(-3)|$Area $= \frac{1}{2} |0 + 0 - 9|$Area $= \frac{1}{2} |-9| = \frac{9}{2} = 4.5$ square units.

The area of a triangle having vertices $A(3, 0)$,$B(0, 3)$,and $C(3, 3)$ is:

Similar Questions

The coordinates of the midpoint of a line segment joining $A (3, -2)$ and $B (-1, 4)$ are:

The fourth vertex $D$ of a parallelogram $ABCD$ whose three vertices are $A(-2,3)$,$B(6,7)$,and $C(8,3)$ is

State whether the following statements are true or false. Justify your answer.
Points $A(4,3), B(6,4), C(5,-6)$ and $D(-3,5)$ are the vertices of a parallelogram.

$A (h, k), B (1, 1)$ and $C (2, 1)$ are the vertices of $\Delta ABC$. If the area of $\Delta ABC$ is $1$,then find the possible values of $k$.

The perimeter of a triangle with vertices $(0,4), (0,0)$ and $(3,0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module