156,716 views
29 votes
29 votes
A(3,8), B(3,-2), C( t ,1 ), if AB=2BC, find the possible values of t.​

User Bonna
by
2.9k points

1 Answer

23 votes
23 votes

Find the distance between A and B, AB,


AB=√((3-3)^2+(-2-8)^2)=10

Next, find the distance between B and C, BC,


BC=√((t-3)^2+(1-2)^2)=√((t-3)^2+1)

The equation is,


10=2√((t-3)^2+1)


25=(t-3)^2+1


t^2-3t+10=25


t^2-3t-15=0

Use quadratic formula to get possible values of t,


t_1=(3+√(69))/(2)


t_2=(3-√(69))/(2)

Hope this helps :)

User Robynhenderson
by
2.7k points