Linear Algebra and Its Applications, 4th Edition

Published by Brooks Cole
ISBN 10: 0030105676
ISBN 13: 978-0-03010-567-8

Chapter 1 - Section 1.2 - The Geometry of Linear Equations - Problem Set: 10

Answer

$y_1 = a*(y_2 - (y_3/2))$ where a is a real number

Work Step by Step

For all the three points to lie on a straight line, one point must be a linear combination of the other two. This implies $(0,y_1) = a\times(1,y_2)+b\times(2,y_2)$ Equating x-coordinates on both sides gives the following relation : $0 = a + 2*b$ Therefore, $b = -a/2$ Now equate the y-coordinates on both sides of the equation, $y_1 = a*y_2 + b*y_3$ substituting $b = -a/2$ in the above equation we get $y_1 = a*(y_2 - (y_3/2))$ where a can be any real number
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