Question

Find the area of triangle whose vertices are A ( 1,2 ),B ( 3,4 ), C ( 1/2,1/4 )

Answer 1

Answer:

put the formula

area of tiangle of co ordinate geometry

Answer 2

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

Given that vertices of triangle ABC are

$\purple{\rm :\longmapsto\:(x_1,y_1) = (1,2)}$

$\purple{\rm :\longmapsto\:(x_2,y_2) = (3,4)}$

$\purple{\rm :\longmapsto\:(x_3,y_3) = \bigg(\dfrac{1}{2} ,\dfrac{1}{4} \bigg)}$

We know,

➢ Area of triangle is given by

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

So, on substituting the values of

$\purple{\rm :\longmapsto\:x_1 = 1}$

$\purple{\rm :\longmapsto\:x_2 = 3}$

$\purple{\rm :\longmapsto\:x_3 = \dfrac{1}{2} }$

$\purple{\rm :\longmapsto\:y_1 = 2}$

$\purple{\rm :\longmapsto\:y_2 = 4}$

$\purple{\rm :\longmapsto\:y_3 = \dfrac{1}{4} }$

we get

$\rm \ Area =\dfrac{1}{2}\bigg |1\bigg(4 - \dfrac{1}{4} \bigg) + 3\bigg(\dfrac{1}{4} - 2\bigg) + \dfrac{1}{2}(2 - 4) \bigg|$

$\rm \ Area =\dfrac{1}{2}\bigg |\bigg(\dfrac{16 - 1}{4} \bigg) + 3\bigg(\dfrac{1 - 8}{4}\bigg) + \dfrac{1}{2}( - 2) \bigg|$

$\rm \ Area =\dfrac{1}{2}\bigg |\dfrac{15}{4} - \dfrac{21}{4} - 1 \bigg|$

$\rm \ Area =\dfrac{1}{2}\bigg |\dfrac{15 - 21 - 4}{4} \bigg|$

$\rm \ Area =\dfrac{1}{2}\bigg |\dfrac{ - 10}{4} \bigg|$

$\rm\implies \:\bf \ Area =\dfrac{5}{4} \: square \: units$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Learn More:

1. Section formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\sf\implies \boxed{\tt{ (x,y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}}$

2. Mid-point formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\sf\implies \boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}}$

3. Centroid of a triangle

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\sf\implies \boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}}$