Question

Find a relation between x and y such that the point (x,y) is equidistant from
(5,1) and (-1,5).

Answer 1

Answer:

Explanation:

Given :

• Points are A(5 , 1) and B(-1 , 5).

To Find :

• A relation between x and y.

Solution :

Let, P(x, y) be equidistant from the points

A(5, 1) and B(-1, 5).

.°. AP = PB

=> √(x - 5)² + (y - 1)² = √(x + 1)² + (y - 5)²

[ Squaring on both sides, we get ]

=> (x - 5)² + (y - 1)² = (x + 1)² + (y - 5)²

=> x² - 10x + 25 + y² - 2y + 1 = x² + 2x + 1 + y² - 10y + 25

=> -10x - 2y + 26 = 2x - 10y + 26

=> -10x - 2y = 2x - 10y

=> -10x - 2x - 2y + 10y = 0

=> -12x + 8y = 0

=> 12x - 8y = 0

=> 3x - 2y = 0

Hence :

A relation between x and y is 3x - 2y = 0.

Answer 2

$\huge\red{\boxed{\blue{\mathtt{\overbrace{\underbrace{\fcolorbox{red}{black}{\underline{\red{★ANSWER★}}}}}}}}}$

$\red{GIVEN :}$

$\blue\bigstar$Points that are given = (5, 1) and (-1, 5).

$\red{TO \: FIND :}$

$\blue\bigstar$Find a relation between x and y.

$\red{SOLUTION :}$

$\blue\bigstar$Let points(x, y) equidistant from (5, -1) and (-1, 5).

➔ √(x - 5)² + (y - 1)² = √(x + 1)² + (y - 5)²

After squaring on both the sides :

➔ (x - 5)² + (y - 1)² = (x + 1)² + (y - 5)²

➔ x² - 10x + 25 + y² - 2y + 1 = x² + 2x + 1 + y² - 10y + 25

➔ -10x - 2y + 26 = 2x - 10y + 26

➔ -10x - 2y = 2x - 10y

➔ -10x - 2x - 2y + 10y = 0

➔ -12x + 8y = 0

➔ 12x - 8y = 0

➔ 3x - 2y = 0

$\small\boxed{{\mathtt\red{Hence, \: A \: relation \: between \: x \: and \: y \: is \: 3x - 2y = 0}}}$

Hope it Helps Buddy ♥️