Question

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

(6,5) and (-4,3).

जिन्दु(x,y), जिन्दओु ं (6,5) और (-4,3) सेबर बर दरूी पर है| x तथ y के बीच सबं धं जिरूजपत करें|​

Answer 1

Given : point (x,y) is equidistant from (6,5) and (-4,3).

To find : relation between x and y

Solution:

point (x,y) is equidistant from (6,5) and (-4,3).

=> √(x - 6)² + (y - 5)² = √(x -(-4))² + (y - 3)²

=> (x - 6)² + (y - 5)² =  (x + 4)² + (y - 3)²

=> x² - 12x + 36+ y² -10y + 25  = x² + 8x + 16  + y² - 6y + 9

=>  -20x - 4y  + 36 = 0

=> -5x - y + 9 = 0

=> 5x + y = 9

Another way :

Slope of (6,5 ) and (-4 , 3)

= 1/5

sloe of perpendicular line = - 5

Mid point of (6,5 ) and (-4 , 3) = (1 , 4)

Equation of line = y - 4 = (-5)(x -1)

=> y - 4 = -5x + 5

=> 5x + y = 9

Learn more:

Find the value of x if p(x 2) is equidistant from the point a(2 -1)and b ...

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The point which is equidistant to the points (9, 3),(7, -1) ,(-1,3)

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Answer 2

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5x + y = 9