[tex]\boxed{\sf \: A(-1,-2) , B(3,-5) , C(8,-2), D(3, - 2) \: are \: not \: the \: vertices \: of \: a \: parallelogram. \: }\\ \\ [/tex]
Step-by-step explanation:
We know,
In parallelogram, diagonals bisect each other.
So in order to prove that given vertices of A, B, C, D taken in order forms a parallelogram, it is sufficient to show that midpoint of AC is equals to midpoint of BD.
Given coordinates are A(- 1, - 2) , B(3, - 5) , C(8, - 2) and D(3, - 2)
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
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
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
Answers & Comments
Verified answer
Answer:
[tex]\boxed{\sf \: A(-1,-2) , B(3,-5) , C(8,-2), D(3, - 2) \: are \: not \: the \: vertices \: of \: a \: parallelogram. \: }\\ \\ [/tex]
Step-by-step explanation:
We know,
In parallelogram, diagonals bisect each other.
So in order to prove that given vertices of A, B, C, D taken in order forms a parallelogram, it is sufficient to show that midpoint of AC is equals to midpoint of BD.
Given coordinates are A(- 1, - 2) , B(3, - 5) , C(8, - 2) and D(3, - 2)
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
[tex]\begin{gathered}\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}} \\ \end{gathered} \\ [/tex]
So, Let us first find midpoint of AC.
Coordinates of A = ( - 1, - 2)
Coordinates of C = (8, - 2)
So, Using midpoint Formula, we get
[tex]\sf \: Midpoint \: of \: AC = \bigg(\dfrac{ - 1 + 8}{2} ,\dfrac{ - 2 - 2}{2} \bigg) \\ \\ [/tex]
[tex]\sf \: Midpoint \: of \: AC = \bigg(\dfrac{7}{2} ,\dfrac{ - 4}{2} \bigg) \\ \\ [/tex]
[tex]\implies\sf \: \sf \: Midpoint \: of \: AC = \bigg(\dfrac{7}{2} , - 2 \bigg) - - - (1) \\ \\ [/tex]
Now, Lets find the midpoint of BD,
Coordinates of B = (3, - 5)
Coordinates of D = (3, - 2)
So, Using Midpoint Formula,
[tex]\sf \: Midpoint \: of \: BD = \bigg(\dfrac{3 + 3}{2} ,\dfrac{ - 5 - 2}{2} \bigg) \\ \\ [/tex]
[tex]\sf \: Midpoint \: of \: BD = \bigg(\dfrac{6}{2} ,\dfrac{ - 7}{2} \bigg) \\ \\ [/tex]
[tex]\implies\sf \: \sf \: Midpoint \: of \: BD = \bigg(3 ,\dfrac{ - 7}{2} \bigg) - - - (2)\\ \\ [/tex]
From equation (1) and (2), we concluded that
[tex]\implies\sf \: Midpoint \: of \: AC \: \ne \: Midpoint \: of \: BD \\ \\ [/tex]
[tex]\sf\implies \:A(-1,-2) , B(3,-5) , C(8,-2), D(3, - 2) \: are \: not \: the \: vertices \: of \: a \: parallelogram. \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] { \large{ \sf{Additional\:Information}}}[/tex]
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
[tex]\begin{gathered}\boxed{\tt{ AB \: = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} }}} \\ \end{gathered} \\ [/tex]
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
[tex]\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} \\ [/tex]
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
[tex]\begin{gathered}\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}} \\ \end{gathered} \\ [/tex]
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
[tex]\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} \\ [/tex]
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
[tex]\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} \\ [/tex]