Chapter 6.3, Problem 71E

Solution

a)

To prove: the point $P\left(x,y\right)$ on $C$ is equal to $P_1\left(a,b\right)$ when $t=0$ .

It is proved that the point $P\left(x,y\right)$ on $C$ is equal to $P_1\left(a,b\right)$ when $t=0$ .

Given information:

The parametric equations: $x=tc+\left(1-t\right)a$ , $y=td+\left(1-t\right)b$ where $P_1\left(a,b\right)$ and $P_2\left(c,d\right)$ are two fixed points.

Calculation:

Substitute 0 for $t$ in the equations: $x=tc+\left(1-t\right)a$ , $y=td+\left(1-t\right)b$

  $\begin{array}{c}\left(x,y\right)=\left(tc+\left(1-t\right)a,td+\left(1-t\right)b\right)\\ =\left(\left(0\right)c+\left(1-0\right)a,\left(0\right)d+\left(1-0\right)b\right)\\ =\left(0+a,0+b\right)\\ =\left(a,b\right)\end{array}$

Thus, the point $P\left(x,y\right)$ on $C$ is equal to $P_1\left(a,b\right)$ when $t=0$ .

b)

To prove: the point $P\left(x,y\right)$ on $C$ is equal to $P_1\left(a,b\right)$ when $t=0$ .

It is proved that the point $P\left(x,y\right)$ on $C$ is equal to $P_2\left(c,d\right)$ when $t=1$ .

Given information:

The parametric equations: $x=tc+\left(1-t\right)a$ , $y=td+\left(1-t\right)b$ where $P_1\left(a,b\right)$ and $P_2\left(c,d\right)$ are two fixed points.

Calculation:

Substitute 1 for $t$ in the equations: $x=tc+\left(1-t\right)a$ , $y=td+\left(1-t\right)b$

  $\begin{array}{c}\left(x,y\right)=\left(tc+\left(1-t\right)a,td+\left(1-t\right)b\right)\\ =\left(\left(1\right)c+\left(1-1\right)a,\left(1\right)d+\left(1-1\right)b\right)\\ =\left(c+\left(0\right)a,d+\left(0\right)b\right)\\ =\left(c,d\right)\end{array}$

Thus, the point $P\left(x,y\right)$ on $C$ is equal to $P_2\left(c,d\right)$ when $t=1$ .