>
Find the value of k, if the point P 0, 2 equidistant from 3, k and k , 5.
Solution
P(0,2) Q(3,k) R(k,5)
PQ =
PR =
Hence k =1 for P to be equidistant from Q and R
Mathematics
RD Sharma
Standard X
3
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)`
It is said that P(0,2) is equidistant from both A(3,k) and B(k,5).
So, using the distance formula for both these pairs of points we have
`AP =sqrt((3)^2 + (k - 2)^2)`
`BP = sqrt((k)^2 + (3)^2)`
Now since both these distances are given to be the same, let us equate both.
AP = Bp
`sqrt((3)^2 + (k -2)^2) = sqrt((k)^2 + (3)^2)`
Squaring on both sides we have,
`(3)^2 + (k - 2)^2 = (k)^2 + (3)^2`
`9 + k^2 + 4 - 4k = k^2 + 9`
4k = 4
k = 1
Hence the value of ‘k’ for which the point ‘P’ is equidistant from the other two given points is k = 1