Question

find the area of the triangle formed by the points V(0,5), S(0,2) and T(3,0)​

Answer 1

Given :-

The vertices of triangle VST

• V(0, 5)

• S(0, 2)

• T(3, 0)

To Find :-

• Area of triangle VST

Formula Used :-

Area (A) of triangle with given vertices is

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

$\large\underline{\sf{Solution-}}$

Given vertices of triangle VST,

• V(0, 5)

• S(0, 2)

• T(3, 0)

We know,

• Area of triangle is

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

Here,

• • x₁ = 0

• • x₂ = 0

• • x₃ = 3

• • y₁ = 5

• • y₂ = 2

• • y₃ = 0

On substituting all these values in formula,

$\rm :\longmapsto\:\sf \ Area=\dfrac{1}{2} \bigg|0(2 - 0) + 0(0 - 5) + 3(5 - 2) \bigg|$

$\rm :\longmapsto\:\sf \ Area=\dfrac{1}{2} \bigg| 3 \times 3 \bigg|$

$\rm :\longmapsto\:\sf \ Area=\dfrac{9}{2} \: square \: units$

Additional Information :-

1. Distance between two points is calculated by using the formula given below,

$\rm D = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} }$

2. Section Formula is used to find the co ordinates of the point(Q) which divides the line segment joining the points (B) and (C) internally

${\underline{\boxed{\sf{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} , \: \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$