Question

If the distance between the point (1,0) and (4,k) is 5 then what can be the possible values of k?

Answer 1

by mid point theorem

x1+x2/2 +y1+y2/2=5

1+4/2+0+k/2=5

5/2+k/2=5

k/2=5 - 5/2

k/2=10/2-5/2

k/2=5/2

k=5/2×2

k=5

Answer 2

Answer:

HELLO Mate !

Step-by-step explanation:

given:-

A(1,0) ---> X1 , Y1

B(4,k) ---> X2, Y2

Distance between A & B is 5

answer in the attachment!  

slight correction

:-

5 = 3+ k

k = 2

K ≠ 3/2

T-H-A-N-K-S!