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
Question :- Find the relationship between x and y so that the points (x, y) is equidistant from (7, 1) and (3, 5)
Answer:
[tex]\boxed{ \bf{ \: \qquad \: x \: - \: y \: = \: 2\qquad \:}} \\ \\ [/tex]
Step-by-step explanation:
Let assume the coordinates (x, y), (7, 1) and (3, 5) are represented as P, A, B respectively.
According to statement, it is given that
[tex]\sf \: AP = PB \\ \\ [/tex]
On squaring both sides, we get
[tex]\sf \: AP^{2} = PB^{2} \\ \\ [/tex]
We know,
So, using Distance Formula, we get
[tex]\sf \: {(x - 7)}^{2} + (y - 1)^{2} = (x - 3)^{2} + (y - 5)^{2} \\ \\ [/tex]
[tex]\sf \: {(x - 7)}^{2} - (x - 3)^{2} = (y - 5)^{2} - (y - 1)^{2}\\ \\ [/tex]
[tex]\sf \: (x - 7 + x - 3)(x - 7 - x + 3) = (y - 5 + y - 1)(y - 5 - y + 1) \\ \\ [/tex]
[tex]\sf \: (2x - 10 )( - 4) = (2y -6)( - 4) \\ \\ [/tex]
[tex]\sf \: 2x - 10 = 2y -6 \\ \\ [/tex]
[tex]\sf \: 2x -2y = -6 + 10 \\ \\ [/tex]
[tex]\sf \: 2x -2y = 4 \\ \\ [/tex]
[tex]\sf \: 2(x -y) = 4 \\ \\ [/tex]
[tex]\bf\implies \:x - y = 2 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Basic Formulae Used
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{\sf{ AB \: = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} }}} \\ \end{gathered} \\ [/tex]
2. Algebraic Identity
[tex]\boxed{ \sf{ \: {x}^{2} - {y}^{2} = (x - y)(x + y) \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{ \bf{Additional\:Information}}}[/tex]
1. 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]
2. 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]
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
[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]
4. 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]
Verified answer
[tex]\underline{\underline{\bf{Solution : -}}}[/tex]
Let assume the coordinates (x, y), (7, 1) and (3, 5) are represented as P, A, B respectively.
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\implies{PA² = PB²}[/tex]
[tex]\sf\implies{(x - 7)² + (y - 1)² = (x - 3)² + (y - 5)² }[/tex]
By using identity :
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\underline{\underline{\red{\bf{(x - y)² = x² + y² - 2xy }}}}[/tex]
[tex]\sf\implies{x² + 49 - 14x + y² + 1 -2y = x² +9 - 6x + y² + 25y - 10y}[/tex]
[tex]\sf\implies{50 - 14x - 2y - 10y + 34 - 50 = 0}[/tex]
[tex]\sf\implies{8x - 8y - 16 = 0 }[/tex]
[tex]\sf\implies{8 (x - y - 2) = 0}[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\bf\implies{x - y = 2}[/tex]
The difference of x and y is 2.
[tex]\rule{200pt}{4pt}[/tex]