Question

The mid point of sides of triangleABC. With vertices are A(1, -1), B(-4, 6), C(-3,-5) are D, E, F rispectively. Find the area of triangle DEF.​

Answer 1

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

Given that,

• Coordinates of A = (1, - 1)

• Coordinates of B = (- 4, 6)

• Coordinates of C = (- 3, - 5)

Let assume that

• D, E, F are the midpoints of AB, BC, CA respectively.

Now, D is the midpoint of AB.

We know, Mid-point formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the mid-point of AB, then the coordinates of C is given by

$\begin{gathered}\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}} \\ \end{gathered}$

So, using Midpoint Formula,

$\rm \: Coordinates\:of\:D = \bigg(\dfrac{1 - 4}{2}, \: \dfrac{ - 1 + 6}{2} \bigg)$

$\rm\implies \: Coordinates\:of\:D = \bigg( - \dfrac{3}{2}, \: \dfrac{5}{2} \bigg)$

Now, E is the midpoint of BC.

So, using Midpoint Formula,

$\rm \: Coordinates\:of\:E = \bigg(\dfrac{- 4 - 3}{2}, \: \dfrac{6 - 5}{2} \bigg)$

$\rm\implies \: Coordinates\:of\:E = \bigg( - \dfrac{7}{2}, \: \dfrac{1}{2} \bigg)$

Now, F is the midpoint of AC.

So, using Midpoint Formula,

$\rm \: Coordinates\:of\:F = \bigg(\dfrac{1 - 3}{2}, \: \dfrac{ - 1 - 5}{2} \bigg)$

$\rm \: Coordinates\:of\:F = \bigg(\dfrac{- 2}{2}, \: \dfrac{ -6}{2} \bigg)$

$\rm\implies \:Coordinates\:of\:F = ( - 1, \: - 3)$

Now, We know Area of a triangle is evaluated as

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle, then the area of triangle is given by

$\begin{gathered}\boxed{\rm{ Area =\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|}} \\ \end{gathered}$

Using the above result

$\rm \: Area\: of\:\triangle\:DEF$

$\rm \: = \dfrac{1}{2}\bigg | - \dfrac{3}{2} \bigg(\dfrac{1}{2}+ 3\bigg) - \dfrac{7}{2} \bigg( - 3 - \dfrac{5}{2} \bigg) - 1\bigg(\dfrac{5}{2} - \dfrac{1}{2} \bigg) \bigg|$

$\rm \: = \dfrac{1}{2}\bigg | - \dfrac{3}{2} \bigg(\dfrac{7}{2}\bigg) - \dfrac{7}{2} \bigg( - \dfrac{11}{2} \bigg) - 1\bigg(\dfrac{5 - 1}{2}\bigg) \bigg|$

$\rm \: = \dfrac{1}{2}\bigg | - \dfrac{21}{4} + \dfrac{77}{4} - 2 \bigg|$

$\rm \: = \dfrac{1}{2}\bigg | \dfrac{ - 21 + 77}{4} - 2 \bigg|$

$\rm \: = \dfrac{1}{2}\bigg | \dfrac{56}{4} - 2 \bigg|$

$\rm \: = \dfrac{1}{2}\bigg | 14 - 2 \bigg|$

$\rm \: = \dfrac{1}{2} \times 12$

$\rm \: = \: 6 \: square \: units$

Hence,

$\rm\implies \:\bf \: Area\: of\:\triangle\:DEF = 6 \: square \: units$

$\rule{190pt}{2pt}$

Additional Information :-

1. Distance Formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane, then distance between A and B is given by

$\begin{gathered}\boxed{\tt{ \: \: AB \: = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} } \: \: \: }} \\ \end{gathered}$

2. Section formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂, then the coordinates of C is given by

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

3. Centroid of a triangle

Centroid of a triangle is defined as the point at which the medians of the triangle meet and is represented by the symbol G.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle and G(x, y) be the centroid of the triangle, then the coordinates of G is given by

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