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Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (− 3, 4). - Mathematics

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Find a relation between x and y such that the point (xy) is equidistant from the point (3, 6) and (− 3, 4).

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Solution

Point (xy) is equidistant from (3, 6) and (−3, 4).

`:.sqrt((x-3)^2+(y-6)^2) = sqrt((x-(-3))^2 + (y -4)^2)`

`sqrt((x-3)^2+(y-6)^2)=sqrt((x+3)^2+(y-4)^2)`

`(x-3)^2 + (y -6)^2 = (x+3)^2 + (y-4)^2`

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

36 - 16 = 6x + 6x + 12y - 8y

20 = 12x + 4y

3x + y = 5

3x + y - 5 = 0

Concept: Distance Formula
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APPEARS IN

NCERT Class 10 Maths
Chapter 7 Coordinate Geometry
Exercise 7.1 | Q 10 | Page 162

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