NCERT Solutions for Class 10 Maths Chapter 7
Important NCERT Questions
Coordinate Geometry
NCERT Books for Session 2022-2023
CBSE Board and UP Board Others state Board
EXERCISE 7.1
Page No:161
Questions No:10
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).
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Point P(x,y) is equidistant from A(3,6) and B(-3,4). Therefore, PA = PB
⇒ √((3-x)²+(6-y)²) = √((-3-x)²)+(4-y)²)
⇒ √(9+x²-6x+36+y²-12y) = √(9+x²+6x+16+y²-8y)
Squaring both sides
9+x²-6x+36+y²-12y = 9+x²+6x+16+y²-8y
⇒ -12x-4y = -20
⇒ 3x+y=5
Here is video explanation (～￣▽￣)～