>> Suppose parametric equations for the line segment between the point 8 comma 5
and the point 2 comma 1 have the form X equals A plus BT and Y equals C plus DT.
If the parametric curve, or in this case segment, starts at the point 8 comma 5,
when T equals zero, which would be this point here, and ends at the point 2 comma 1,
when T equals 1, which would be this point here, we want to find ABC
and D. Before we do this though, notice how as T increases, the segment will be traced
in this direction here, which is called the orientation of the curve,
or in this case the orientation of the segment.
Because we have to be at the point 8 comma 5 when T equals zero,
this means X of zero must equal the S coordinate 8 and Y of zero must equal the Y coordinate 5.
And when T equals 1, we must be at this point,
which means X of 1 must equal 2, and Y of 1 must equal 1.
Let's begin by determining X of zero and Y of zero.
X of zero would be equal to A plus B times T,
but we know T is zero, so we'd have B times zero.
Again, must be equal to the X coordinate when T is zero, which is equal to 8.
Notice how this tells us that A must be equal to 8.
Now let's find Y of zero, which we know must equal 5.
Y of zero must equal C plus D times T, which is zero, and this must equal 5.
So this tells us that C equals 5.
So now that we know A equals 8 and C equals 5, let's find X or 1 and Y of 1 where X
of 1 must equal 2 and Y of 1 must equal 1.
So X of 1 would be equal to A plus B times T, but T is 1, so we have B times 1.
Again, this must equal 2.
But we also know that A equals 8, so this gives us the equation 8 plus B equals 2 subtracting 8
on both sides, notice how B equals negative 6.
And now we'll find Y of 1, which we know must equal 1.
Y of 1 equals C plus D times 1, which must equal 1.
And we know C is 5, so 5 plus D equals 1 subtracting 5
on both sides, D equals negative 4.
So now we have the values of ABC and D where our parametric equations would be X 1X
of T equals A plus BT, which would be 8 minus 6T and Y
or Y of T would be equal to C plus DT or 5 minus 4T.
I do want to point out one more thing before we go though.
Notice how the equations for X and T are linear, mean they do fit the form of Y equals MX plus B,
where in this form M is the slope, which tells us the change of Y with respect to X.
But in this case B would be the change of X with respect to T and D would be the change of Y
with respect to T. So we could have found B, which again would be the change of X divided
by the change of T by finding the change of the X coordinates and divide by the change
of T. Notice how the change of X would be 2 minus 8 and the change
of T would be 1 minus zero, which equals negative 6,
which again is a value of B that we found.
We could also have found D by determining the change of Y divided by the change of T,
where the change of Y would be 1 minus 5 and the change of T would be 1 minus zero,
giving us negative 4 for D, which again we already found.
But I think it is important to recognize that the values of B and D do represent the change
of X with respect to T and the change of Y with respect to T in our parametric equations.
I hope you found this helpful. ------------------------------9aea49991091--