Question

17. If the point P(k - 1, 2) is equidistant from the points A
(3,k) and B(k, 5), find the value of k.​

Answer 1

Rule :

If we take two points (x₁, y₁) and (x₂, y₂), then the distance between the two points is given by

= √{(x₁ - x₂)² + (y₁ - y₂)²} units

Solution :

The distance between the points P (k - 1, 2) and A (3, k) is

|AP| = √{(k - 1 - 3)² + (2 - k)²} units

= √{(k - 4)² + (k - 2)²} units

= √(k² - 8k + 16 + k² - 4k + 4) units

= √(2k² - 12k + 20) units

and the distance between the points P (k - 1, 2) and B (k, 5) is

|BP| = √{(k - 1 - k)² + (2 - 5)²} units

= √(1 + 9) units

= √10 units

ATQ,

|AP| = |BP|

or, |AP|² = |BP|²

or, 2k² - 12k + 20 = 10

or, 2k² - 12k + 10 = 0

or, k² - 6k + 5 = 0

or, (k - 5) (k - 1) = 0

Either k - 5 = 0 or, k - 1 = 0

i.e., k = 5, 1

Therefore, the required value of k is

k = 5, 1