Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 10 - Section 10.5 - Equations of Lines - Exercises - Page 472: 44


$(0, ((-a^2/b)+b)$

Work Step by Step

Based on the definition of an isosceles triangle, we know that the two lines will intersect at x=0. The slope of NQ, which is perpendicular to the bisector line, is given by -2b/2a. Thus, we know that the slope of the perpendicular bisector line is: a/b We know that a point on the line is the midpoint of N and Q. Thus, we know: $M= (\frac{0+2a}{2}, \frac{0+2b}{2}) = (a,b)$ This means that the line is: $y-b = (a/b)(x-a)$ Using the same reasoning, the slope of the other perpendicular bisector is: -a/b And symmetry tells us that the equation of the other line is: $y-b = (-a/b)(x+a)$ Here, we know that the intersection point must occur at x=0. Thus, we have the equation: $y-b = (-a/b)(x+a)$ $y-b = (-a/b)(a)$ $y = (-a^2/b)+b$ This gives the answer: $(0, ((-a^2/b)+b)$
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