## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.5 - Parametric Equations - Exercise Set - Page 1020: 52

#### Answer

$x=6t+3$ ; $y=13t-1$

#### Work Step by Step

For a line, we have: $(3,-1); (9,12)$ $x_1=3, x_2=9,y_1=-1, y_2=12$ Now, in parametric from: $x=(x_2-x_1) t+x_1=6t+3$ and $y=(y_2-y_1)t_1+y_1 =13t-1$ So, $x=6t+3$ ; $y=13t-1$

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