Question

Find area of the triangle with vertices at the point given in each of the following (-2, -3), (3, 2), (-1, -8)

Answer 1

Answer:

We have the formula to find area of triangle whose all vertices are given

as

(x1, y1) , (x2, y2) , (x3, y3)

i.e.

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

Now,

using above formula we can easily find the area of triangle whose vertices are given in the question as:-)

$Area = \frac{1}{2} | - 2(2 + 8) + 3( - 8 + 3) + ( - 1)( - 3 - 2) | \\ \\ = \frac{1}{2} | - 20 - 15 + 5| \\ \\ = \frac{1}{2} | - 30| \\ \\ = \frac{1}{2} \times 30 \\ \\ = 15sq. \: units$

Answer 2

AnswEr :

Let's know the rule for finding Area of Triangle with given Vertices.

let's assume Vertices of Triangle as :

⋆ (x₁ , y₁) ; (x₂ , y₂) and (x₃3 , y₃)

$\boxed{ \small\sf{Ar. \: of \: \triangle= \dfrac{1}{2} \times | x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2 )| }}$

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• Let's Head to the Question Now :

⋆ Given Vertices Here is (-2, -3), (3, 2), (-1, -8)

$\small\sf{Ar. \: of \: \triangle= \dfrac{1}{2} \times | x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|}$

$\small\sf{Ar. \: of \: \triangle= \dfrac{1}{2} \times | - 2(2-( - 8))+3( - 8 - ( - 3))+( - 1)( - 3-2)|}$

$\small\sf{Ar. \: of \: \triangle= \dfrac{1}{2} \times | - 2(2 + 8)+3( - 8 + 3) - 1( - 3-2)|}$

$\small\sf{Ar. \: of \: \triangle= \dfrac{1}{2} \times | (- 2 \times 10)+(3 \times - 5) - (1 \times - 5)|}$

$\small\sf{Ar. \: of \: \triangle= \dfrac{1}{2} \times | - 20+( - 15) - (- 5)|}$

$\sf{Ar. \: of \: \triangle= \dfrac{1}{2} \times | - 20 - 15 + 5|}$

$\sf{Ar. \: of \: \triangle= \dfrac{1}{2} \times |- 35 + 5|}$

$\sf{Ar. \: of \: \triangle= \dfrac{1}{2} \times | - 30|}$

$\sf{Ar. \: of \: \triangle= \dfrac{1}{ \cancel2} \times \cancel{30}}$

$\large\boxed{\sf{Ar. \: of \: \triangle= 15}}$

⠀

჻ Hence, Area of $\triangle$ is 15 square units.