Jon,
The statement above makes sense but like you I don't see how that relates to 1.2.&3??
Sorry seems like something is missing.
Jim
Jim S.
tutor
Jon,
Ok now i see what's required. You were give the parametric equation for a line between two points: x=x1+(x2-x1)*t and y=y1+(y2-y1)*t where t>=0 and <=1. This is a general form for any two points (x1,y1) and (x2,y2). To construct specific equation
for the line going from A to B you let x1=1,y1=1 and x2=4, y2=3 and substitute in the general equation. So the answer to 1. is x=1+3*t, y=1+2*t. You can see when t=0 x=1 and y=1 i.e. A and when t=1 x=4 and y=3 which is point B.
You try doing B to C and A to C
Good luck
Jim
Report
01/08/15
Jon P.
"and P2(x2,y2). Draw the triangle with vertices A(1, 1), B(4, 3), C(1, 7). Find the parametrization, including endpoints, and sketch to check. (Enter your answers as a comma-separated list of equations. Let x and y be in terms of t.)"
01/07/15