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Get Full Access to Linear Algebra: A Geometric Approach - 2 Edition - Chapter 5.2 - Problem 14
Get Full Access to Linear Algebra: A Geometric Approach - 2 Edition - Chapter 5.2 - Problem 14

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# As we saw in Exercises 1.6.7 and 1.6.11, through any three noncollinear points in R2

ISBN: 9781429215213 438

## Solution for problem 14 Chapter 5.2

Linear Algebra: A Geometric Approach | 2nd Edition

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Problem 14

As we saw in Exercises 1.6.7 and 1.6.11, through any three noncollinear points in R2 there pass a unique parabola y = ax2 + bx + c and a unique circle x2 + y2 + ax + by + c = 0. (In the case of the parabola, we must also assume that no two of the points lie on a vertical line.) Given three such points (x1, y1), (x2, y2), and (x3, y3), show that the equation of the parabola and circle are, respectively, det 1 x x2 y 1 x1 x2 1 y1 1 x2 x2 2 y2 1 x3 x2 3 y3 = 0 and det 1 x y x 2 + y2 1 x1 y1 x2 1 + y2 1 1 x2 y2 x2 2 + y2 2 1 x3 y3 x2 3 + y2 3 = 0.

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Mahjabin Haque Homework 01 Algorithm Analysis (Chapter 2) Due 2016/06/02 at 5:00 PM Answer the following questions from the textbook: 2.1, 2.7abc, 2.8bcde, 2.11abcd, 2.12abcd, 2.14, 2.25abcd, 2.31 2.1 Answer: Growth rate order: 2/N, 37, N , N, N log log N, N log N, N log (N ), N log N, N , N , N 2 1.5 2 2 3 n/2 N log N, N , 2 , 2 N log N and N log N are the two functions that grow at the same rate. 2.7 Answer: a. i. O(N) 2 ii. O(N ) iii. O(N ) iv. O(N ) 5 v

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##### ISBN: 9781429215213

This textbook survival guide was created for the textbook: Linear Algebra: A Geometric Approach, edition: 2. This full solution covers the following key subjects: . This expansive textbook survival guide covers 31 chapters, and 547 solutions. The full step-by-step solution to problem: 14 from chapter: 5.2 was answered by , our top Math solution expert on 03/15/18, 05:30PM. Linear Algebra: A Geometric Approach was written by and is associated to the ISBN: 9781429215213. Since the solution to 14 from 5.2 chapter was answered, more than 211 students have viewed the full step-by-step answer. The answer to “As we saw in Exercises 1.6.7 and 1.6.11, through any three noncollinear points in R2 there pass a unique parabola y = ax2 + bx + c and a unique circle x2 + y2 + ax + by + c = 0. (In the case of the parabola, we must also assume that no two of the points lie on a vertical line.) Given three such points (x1, y1), (x2, y2), and (x3, y3), show that the equation of the parabola and circle are, respectively, det 1 x x2 y 1 x1 x2 1 y1 1 x2 x2 2 y2 1 x3 x2 3 y3 = 0 and det 1 x y x 2 + y2 1 x1 y1 x2 1 + y2 1 1 x2 y2 x2 2 + y2 2 1 x3 y3 x2 3 + y2 3 = 0.” is broken down into a number of easy to follow steps, and 142 words.

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