Question

Find the parametric equation of the line r(t), 0 ≤ t ≤ 1, between P = (2, − 3, 1) and Q = (− 2, 1, 7).

Answer 1

Answer:

x=-2+3t , y=-9-10t

Step-by-step explanation:

I'm going to start by solving the first equation for t....which will give (x-x1)/(x2-x1) =t.

You can now substitute this in the second equation to get....y=y1 + (y2-y1)(x-x1)/(x2-x1)

Rearranging, you get y-y1 = (y2-y1) / (x2-x1) * (x-x1) which is equivalent to y-y1= m(x-x1) which is the point-slope formula for a line.

So since t was varying between 0 and 1,

plugging in t=0 you get x=x1 and y = y1 givin you the coordinate (x1,y1)

similarly, plugging in t=1 leads to (x2,y2)

So this is a representation of a segment connecting these 2 points.

B) just plug in the points into your given formula from part a to get...

x=-2 +(1--2)t and y = 9 +(-1-9)t

or x=-2+3t , y=-9-10t