Question

Find the value of k, if the point P(2,4) is equidistant from the points (5,k) and (k,7)

Answer 1

Answer:

k=3

Step-by-step explanation:

Let us take the first point (5,k) as A and the second point (k,7) as B ...We know AP = BP

this implies AP²=BP²-----1

applying distance formula

we know

AP²= (5-2)²+(k-4)²-------2

BP²= (k-2)²+(7-4)²-----3

from 1,2 and 3

3²+(k-4)²=(k-2)²+3²

this implies

(k-2)²-(k-4)²=0

[(k-2)+(k-4)][(k-2)-(k-4)]=0 {a²-b²= (a+b)(a-b)}

[2k-6][2]=0

2k-6=0

k=3