Question

find relation between x and y such that the point (x ,y ) is equidistant from the points (3 , 6) and (-3,4)

Answer 1

apply distance formula

√(y2-y1) whole square +(x2-x1)whole square under root are on whole formula

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption

P(x , y)

A( 3, 6)

B( -3 , 4)

Equidistant from P

AP = BP

Now,

Distance Formula = √(x2 - x1)² + (y2 - y1)²

√[(x - 3)² + (y - 6)²] = √[(x + 3)² + (y - 4)²]

Squaring both sides, we get

[(x - 3)² + (y - 6)²] = √[(x + 3)² + (y - 4)²]

Now,

Using Formula,

(a - b)² = a² + b² - 2ab

[x² + 3² - 2 × x × 3 + y² + 6² - 2 × x × 6] = [x² + 3² + 2 × x × 3 + y² + 4² - 2 × x × 4]

[x² + 9 - 6x + y² + 36 - 12y ] = [x² + 9 + 6x + y² + 16 - 8y]

[x² - 6x + y² + 45 - 12y] = [x² + 6x + y² + 25 - 8y]

x² - x² + y² - y² - 6x - 6x - 12y - 8y = 25 - 45

-12 x - 20y = - 20

-4(3x + 5y) = - 20

(3x + 5y) = - 20/4

(3x + 5y) = 5

(3x + 5y) - 5 = 0

Therefore ,

Relation between x and y :-

(3x + 5y) - 5 = 0