## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 9

#### Answer

$x=\frac{\sqrt 2}{2}(x'-y')$ and $y=\frac{\sqrt 2}{2}(x'+y')$

#### Work Step by Step

Step 1. From the given equation, we have $A=3, B=-10, C=3$. Thus we have $cot2\theta=\frac{A-C}{B}=0$, which gives $2\theta=\frac{\pi}{2}$ and the rotation angle $\theta=\frac{\pi}{4}$ Step 2. Using the transformation formula $x=x'cos\theta-y'sin\theta$ and $y=x'sin\theta+y'cos\theta$, we have $x=x'cos45^\circ-y'sin45^\circ=\frac{\sqrt 2}{2}(x'-y')$ $y=x'sin45^\circ+y'cos45^\circ=\frac{\sqrt 2}{2}(x'+y')$

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