Question

given the coordinates F(-4,-2) G(-2,2) H(4,3) J(2,-1) prove if the quadrilateral is a parallelogram or not

Answer 1

Given :-

The vertices of quadrilateral taken in order such that

Coordinates of F (- 4,- 2)

Coordinates of G (- 2, 2)

Coordinates of H (4, 3)

and

Coordinates of J (2, - 1)

To Prove :-

FGHI is a parallelogram or not.

Concept Used :-

We know,

In parallelogram, diagonals bisect each other.

So in order to prove that given vertices F, G, H, I taken in order forms a parallelogram, it is sufficient to show that midpoint of FH is equals to midpoint of GJ.

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

The vertices of quadrilateral taken in order such that

Coordinates of F (- 4,- 2)

Coordinates of G (- 2, 2)

Coordinates of H (4, 3)

and

Coordinates of J (2, - 1)

We know,

Midpoint Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the midpoint of AB, then coordinates of C is

$\boxed{ \quad\rm \:( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg) \quad}$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: B(x_2,y_2)$

Here,Let us first find midpoint of FH.

Coordinates of F (- 4,- 2)

Coordinates of H (4, 3)

Using midpoint Formula,

x₁ = - 4

x₂ = 4

y₁ = - 2

y₂ = 3

So,

$\rm \: Midpoint \: of \: FH = \bigg(\dfrac{ - 4 + 4}{2} ,\dfrac{ - 2 + 3}{2} \bigg)$

$\rm\implies \:\rm \: Midpoint \: of \: FH = \bigg(0 ,\dfrac{1}{2} \bigg) - - - (1)$

Now, To find Midpoint of GJ

Coordinates of G = (- 2, 2)

Coordinates of J = (2, - 1)

Here,

x₁ = - 2

x₂ = 2

y₁ = 2

y₂ = - 1

Thus,

$\rm \: Midpoint \: of \: GJ = \bigg(\dfrac{ - 2 + 2}{2} ,\dfrac{ - 1 + 2}{2} \bigg)$

$\rm\implies \:\rm \: Midpoint \: of \: GJ = \bigg(0 ,\dfrac{1}{2} \bigg) - - - (2)$

So, from equation (1) and (2), we concluded that

$\rm\implies \:Midpoint \: of \: FH \: = \: Midpoint \: of \: GJ$

$\rm\implies \:F,G,H,J \: are \: the \: vertices \: of \: a \: parallelogram.$

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

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

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

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. 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

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

4. 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

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

5. Area of a triangle

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

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

Answer 2

Parallelogram

A parallelogram is a quadrilateral with two pairs of parallel sides. The opposite sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal.

As we went through the concept about Parallelogram, Now, Let's move on finding the solution for our given question.

We have been given that, F (-4, -2), G (-2, 2), H (4, 3), J (2, -1). And we have been asked to prove whether the quadrilateral is a parallelogram or not.

We saw that, A parallelogram must have two pairs of parallel sides. And So we can proclaim that the midpoint of FH and GJ must be equal, so that it can be a parallelogram.

We know that,

${\boxed{\bullet\:{Midpoint\:=\:({\dfrac{x_1\:+\:x_2}{2}},\:{\dfrac{y_1\:+\:y_2}{2}})}}}$

Firstly, Let's take F and H,

• $\sf{F\:(x_1,\:y_1)\:=\:(-4,\:-2)}$
• $\sf{H\:(x_2,\:y_2)\:=\:(4,\:3)}$

Substituting values in Formula, we get,

$\sf\implies{Midpoint\:=\:({\dfrac{x_1\:+\:x_2}{2}},\:{\dfrac{y_1\:+\:y_2}{2}})}$

$\sf\implies{Midpoint\:=\:({\dfrac{-4\:+\:4}{2}},\:{\dfrac{-2\:+\:3}{2}})}$

$\sf\implies{Midpoint\:=\:({\dfrac{0}{2}},\:{\dfrac{1}{2}})}$

$\sf\implies{Midpoint\:=\:(0,\:{\dfrac{1}{2}})}\:---(1)$

Then, Let's take G and J,

• $\sf{G\:(x_1,\:y_1)\:=\:(-2,\:2)}$
• $\sf{J\:(x_2,\:y_2)\:=\:(2,\:-1)}$

Substituting values in Formula, we get,

$\sf\implies{Midpoint\:=\:({\dfrac{x_1\:+\:x_2}{2}},\:{\dfrac{y_1\:+\:y_2}{2}})}$

$\sf\implies{Midpoint\:=\:({\dfrac{-2\:+\:2}{2}},\:{\dfrac{-1\:+\:2}{2}})}$

$\sf\implies{Midpoint\:=\:({\dfrac{0}{2}},\:{\dfrac{1}{2}})}$

$\sf\implies{Midpoint\:=\:(0,\:{\dfrac{1}{2}})}\:---(2)$

From (1) & (2),

We can conclude that (1) = (2).

Hence, The coordinates F (-4, -2), G (-2, 2), H (4, 3), J (2, -1) forms a parallelogram.