The two vertices of a triangle are (3,5) and (-4,-6) and the coordinates of its centroid are (4,3) then the coordinates of its third vertex are:
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The two vertices of a triangle are (3,5) and (-4,-6) and the coordinates of its centroid are (4,3) then the coordinates of its third vertex are:
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[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
The two vertices of a triangle are (3, 5) and (-4, -6) and the coordinates of its centroid are (4, 3).
Let assume that the coordinates of third vertex be (a, b).
We know,
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]
So, on substituting the values, we get
[tex]\sf \: (4, \: 3) = \bigg(\dfrac{3 - 4 + a}{3}, \: \dfrac{5 - 6 + b}{3}\bigg)\\ \\ [/tex]
[tex]\sf \: (4, \: 3) = \bigg(\dfrac{ - 1 + a}{3}, \: \dfrac{ - 1 + b}{3}\bigg)\\ \\ [/tex]
So, on comparing, we get
[tex]\sf \: \dfrac{ - 1 + a}{3} = 4\: \: \: and \: \: \: \dfrac{ - 1 + b}{3} = 3\\ \\ [/tex]
[tex]\sf \: - 1 + a = 12\: \: \: and \: \: \: - 1 + b = 9\\ \\ [/tex]
[tex]\sf \: a = 12 + 1\: \: \: and \: \: \: b = 9 + 1\\ \\ [/tex]
[tex]\sf \: \bf\implies \: a = 13\: \: \: and \: \: \: b = 10\\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{\bf{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]
Answer:
The co-ordinates of third vertex is (13,10).
Step-by-step explanation:
To Find:
Formula used:
G = [(x1+x2+x3)/3,(y1+y2+y3)/3]
Let the third vertex be (x, y)
Using the above formula, we have
G = [(3-4+x)/3,(5-6+y)/3]
(4,3) = [(x-1)/3,(y-1)/3]
Comparing x and y co-ordinates, we get
(x-1)/3 = 4 and (y-1)/3 = 3
x-1 = 12 and y-1 = 9
x = 13 and y = 10
x,y = 13,10
So, the co-ordinates of third vertex is (13,10).