Question

Find the relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3 , 5)​

Answer 1

Step-by-step explanation:

Given:-

• Two points A(7 , 1) and B(3 , 5)

• The point C(x , y) is equidistant from the points A and B.

To Find:-

• The relation between x and y.

Concept used:-

The distance between two points A(x₁ , y₁) and B(x₂ , y₂)

The distance between AB is

$\sf AB = \sqrt{ (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Solution:-

$\sf A(7 , 1) \: and\: C(x , y)$

$\sf AB = \sqrt{ (x- 7)^{2} + (y - 1)^{2}}$

$\sf \implies AB = \sqrt{ {x}^{2} + 49 - 14x+ {y}^{2} + 1 - 2y }$

$\sf \implies AB = \sqrt{ {x}^{2} + {y}^{2} - 14x - 2y + 50 }$

$\sf B( 3, 5) \: and \: C(x , y)$

$\sf BC = \sqrt{ (x- 3)^{2} + (y - 5)^{2}}$

$\sf \implies BC = \sqrt{ {x}^{2} + 9 - 6x+ {y}^{2} + 25 - 10y }$

$\sf \implies BC = \sqrt{ {x}^{2} + {y}^{2} - 6x - 10y + 34 }$

$\sf As, \: C \: is \: equidistant \: from \: A \: and \: B$

$\sf \implies AC = AB$

$\sf \implies \sqrt{ {x}^{2} + {y}^{2} - 14x - 2y + 50 } = \sqrt{ {x}^{2} + {y}^{2} - 6x - 10y + 34 }$

Squaring on both sides:-

$\sf \implies {x}^{2} + {y}^{2} - 14x - 2y + 50 = {x}^{2} + {y}^{2} - 6x - 10y + 34$

$\sf \implies - 14x - 2y + 50 = - 6x - 10y + 34$

$\sf \implies - 14x + 6x - 2y +10y + 50 - 34= 0$

$\sf \implies - 8x +8y + 16= 0$

Dividing whole equation by 8

$\sf \implies - x +y + 2= 0$

$\sf \implies - x +y= - 2$

$\sf \implies x - y= 2$

Therefore, the relation between x and y is:-

$\large\dashrightarrow \underline{\boxed{\sf x - y= 2}}$

Answer 2

Answer:

$\sf \underline{ \underline{☃Given:}}$

• Two points A(7, 1) and B(3, 5)
• (x , y) is equidistant from the points A and B

$\sf \underline{ \underline{☃Find:}}$

• Relation between x and y

${ \underline {\sf{ \underline{☃Solution:}}}}$

Given, AP = BP

AP² = BP²

By using distance formula ⤵

$\boxed{ \sf{d = \sqrt{ {( x_{2} - x_{1}) }^{2} + {( y_{2} - y_{1})}^{2} } }}$

${\to {\sf{ {(x - 7)}^{2} + {(y - 1)}^{2} = {(x - 3)}^{2} + {(x - 5)}^{2} }}}$

${ \to{ \sf{ {x}^{2} + 49 - 14x + {y}^{2} + 1 - 2y = {x}^{2} + 9 - 6x + {y}^{2} + 25 - 10y}}}$

${ \to{ \sf{ - 7x + 25 - y = - 3x + 17 - 5y}}}$

${ \to{ \sf{ - 4x + 8 + 4y = 0}}}$

${ \to{ \sf{x - y = 2}}}$

Therefore,

The relation between x and y is x-y = 2