Chapter 10.7, Problem 28AYU

$y=-8x+3$

To determine

To find: Two different parametric equations for the rectangular equation.

Answer to Problem 28AYU

$x=t,y=-8t+3$;

$x=t-1,y=-8t+11$

Explanation of Solution

Given:

$y=-8x+3$

Calculation:

Let $x=t$,

$y=-8t+3$

Let $x=t-1$

$y=-8\left(t-1\right)+3$

$y=-8t+8+3$

$y=-8t+11$