howtongchong
howtongchong howtongchong
  • 13-08-2021
  • Mathematics
contestada

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

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r3t40
r3t40 r3t40
  • 13-08-2021

Find the distance between A and B, AB,

[tex]AB=\sqrt{(3-3)^2+(-2-8)^2}=10[/tex]

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

[tex]BC=\sqrt{(t-3)^2+(1-2)^2}=\sqrt{(t-3)^2+1}[/tex]

The equation is,

[tex]10=2\sqrt{(t-3)^2+1}[/tex]

[tex]25=(t-3)^2+1[/tex]

[tex]t^2-3t+10=25[/tex]

[tex]t^2-3t-15=0[/tex]

Use quadratic formula to get possible values of t,

[tex]t_1=\frac{3+\sqrt{69}}{2}[/tex]

[tex]t_2=\frac{3-\sqrt{69}}{2}[/tex]

Hope this helps :)

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