Question

HEEELP :( is this also a possible solution for the problem--
find all values of t such that the points A(t-1, 2t-1), B(4, 1-2t) and C(-3, 3t+5) are collinear.

so y=mx+b so i substituted the x and y values to have

2t-1=m(t-1)+b
1-2t=4m+b
3t+5=-3m+b

but i don't seem to have any success i solving it huhuhuhuhu

Answer 1

well, solve the first two. find m and b in terms of t

Answer 2

@telltoamit so do you think that the system makes sense, though?

Answer 3

yea, t is like a number, say 3

Answer 4

so far my answers don't end up as whole numbers... i've got

t= -25±sqrt(649)
_______________
6

:(

Answer 5

@telltoamit ? what do you think?

Answer 6

no, no

u dint hv to find t

u hv to find m and b

in terms of t

Answer 7

Let's first deal with A and B: A(t-1, 2t-1), B(4, 1-2t)
You correctly stated the equation of a straight line as y=mx+c.
The gradient is calculated by dividing the change in y by the change in x:
m=(1-2t-2t+1)/(4-t+1)=(2-4t)/(5-t)
Now let's deal with A and C:
A(t-1, 2t-1), C(-3, 3t+5)
The gradient between these two is:
m=(3t+5-2t+1)/(-3-t+1)=(t+6)/(-t-2)
So if we equate these two gradients we get:
(t+6)/(-t-2)=2(1-2t)/(5-t)
Multiply to get:
(t+6)(5-t)=-2(1-2t)(t+2)
Expand and simplify to find t. Does this help?

Answer 8

@dydlf. here what Traxter is talking about. it's abvious that the gradient between A and B equals the gradient between B and C also equals the gradient between A and C........mAB=mBC=mAC
..... m=\[\frac{ y_2-y_1 }{ x_2-x_1 }\] u'll find a quadratic equation in (t) solve for t and let us know what u get as answers

Answer 9

i forgot to label my axis sorry for that

Answer 10

Yes I probably should have used a diagram to explain, thanks @hubertH :)