Question

P(6,-1),Q(1,3) and R(a,8) are points and if PQ=QR,then the value of a is ---​

Answer 1

Answer:

PQ^2=(1-6)^2+(3+1)^2=25+16=41

QR^2=(a-1)^2+(8-3)^2=(a-1)^2+25

(a-1)^2+25=41

(a-1)^2=41-25=16

a-1=4, a=5

a-1= -4, a=-3

a= 5 and -3