Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
Solution
The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula
`d = sqrt((x_1-x_2)^2 + (y_1 - y_2)^2)`
Let the three given points be P(x, y), A(3,6) and B(−3,4).
Now let us find the distance between ‘P’ and ‘A’.
`PA = sqrt((x - 3)^2 + (y - 3))`
Now, let us find the distance between ‘P’ and ‘B’.
`PB = sqrt((x + 3)^2 + (y - 6)^2)`
It is given that both these distances are equal. So, let us equate both the above equations,
PA = PB
`sqrt((x - 3)^2 + (y - 6)^2 ) = sqrt((x + 3)^2 + (y - 4)^2)`
Squaring on both sides of the equation we get
`(x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2`
`x^2 + 9 - 6x + y^2 + 36 - 12y = x62 + 9 + 6x + y^2 + 16 - 8y`
12x + 4y = 20
3x + y = 5
Hence the relationship between ‘x’ and ‘y’ based on the given condition is 3x + y = 5
