Question

if Q(3,2) and Q(6,y) are two such points that PQ = 5 find the possible value of y.

Answer 1

Step-by-step explanation:

p(3,2) & Q(6,y)

PQ=5

distance=√(6-3)^2+(y-2)^2

5=√(3)^2+(y^2+4-4y)

5=√9+y^2+4-4y

squaring on both side

25=y^2-4y+13

y^2-4y-12=0

y^2-6y+2y-12=0

y(y-6)+2(y-6)=0

y=6 or y=-2