Question

The distance between two points P(10, 2b) and Q(b. -5) is 5√10. Find the two possible values of b?​

Answer 1

Step-by-step explanation:

We have (-3-b)²+(-2–2b)²=10²

9+b²+6b+4+4b²+8b=100

Or 5b²+14b-87=

I.e. 5b²+29b-15b-87=0

b(5b+29)-3(5b+29)=0

Or (5b+29)(b-3)=0

Now either (5b+29)=0 I.e. b=-29/5

Or (b-3)=0 I.e. b=3

Two possible values of b are -29/5 and 3

Answer 2

this is the solution of your question .