Question

Find the value of k , when the distance between the points (3,2k)and (4,1) is √10 units.

Answer 1

distance between two points = √(x2-x1)² + (y2-y1)²

√10 =√ (4 - 3)² + (1 - 2k)²

⇒ 10 = 1 + 1 + 4k² - 4k

⇒ k² - k -2 = 0

⇒ k² + k - 2k -2 = 0

⇒ k (k + 1) + (-2)(k + 1) =0

⇒ (k -2) (k + 1) = 0

⇒ (k - 2) = 0 or (k + 1) = 0

⇒ k = 2 or k = -1

         HOPE THIS HELPS!!!

Answer 2

Here's your answer:-
see in the pic....

k= 2 or -1

hope this helps u.....