Question

find the value of k for which the distance between the point P(1,2)and Q(-2,k)is 5 units​

Answer 1

using distance fromula
√(x2-x1)²+(y2-y1)²
5= √ (-2-1)²+(k-2)²
square both sides
(5)² = (-3)²+(k-2)²
25 = 9 + k²+4-4k
k²-4k+13-25 =0
k²-4k-12=0
k²-6k+2k-12=0
k(k-6)+2(k-6)=0
(k+2)(k-6)= 0
so,k=-2
k=6

Answer 2

value of K=2 in the following questions