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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). - CBSE Class 10 - Mathematics

#### Question

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

#### 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

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#### APPEARS IN

NCERT Solution for Mathematics Textbook for Class 10 (2019 to Current)
Chapter 7: Coordinate Geometry
Ex. 7.10 | Q: 10 | Page no. 162

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Solution Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (− 3, 4). Concept: Distance Formula.
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