Question

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

Answer 1

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 :)