Find the area of a triangle having the vertic | 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

$3$

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 $A(5, 3)$,$B(4, 5)$,and $C(3, 1)$.Here,$x_1 = 5, y_1 = 3, x_2 = 4, y_2 = 5, x_3 = 3, y_3 = 1$.Substituting these values into the formula:Area $= \frac{1}{2} |5(5 - 1) + 4(1 - 3) + 3(3 - 5)|$Area $= \frac{1}{2} |5(4) + 4(-2) + 3(-2)|$Area $= \frac{1}{2} |20 - 8 - 6|$Area $= \frac{1}{2} |6| = 3 	ext{ square units}$.Thus,the correct option is $B$.

Find the area of a triangle having the vertices $A(5, 3)$,$B(4, 5)$,and $C(3, 1)$.

Similar Questions

If $A (3,0), B (0,0), C (1,3)$ and $D (4,3),$ then $\square ABCD$ is a $\ldots \ldots \ldots \ldots$

State whether the following statement is true or false. Justify your answer.
$A$ circle has its centre at the origin and a point $P(5,0)$ lies on it. The point $Q(6,8)$ lies outside the circle.

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

The vertices of $\Delta ABC$ are $A(3, 4)$,$B(0, 0)$,and $C(6, 0)$. Then the length of the median $\overline{AD}$ is:

Two vertices of a triangle are $A(2, 1)$ and $B(3, -2)$. The third vertex is $C(x, y)$,where $y = x + 3$. If the area of the triangle is $5$,find the coordinates of the third vertex.

Vedclass Test Series

Exam Paper Generator

Online Exam Module