## Calculus 10th Edition

Published by Brooks Cole

# Chapter P - P.2 - Linear Models and Rates of Change - Exercises - Page 17: 72

#### Answer

The points are collinear.

#### Work Step by Step

For three points to be collinear, they must all lie on the same straight line. Therefore, the three given points must satisfy the equation of the same line. To find the equation select any two of the points: $P_{1}$ = ( 0 , (-$a^{2}$ + $b^{2}$ + $c^{2}$)/2c ) $P_{2}$ = ( b/3 , c/3 ) $P_{3}$ = ( b , ($a^{2}$ - $b^{2}$)/c ) Using points $P_{2}$ and $P_{3}$, calculate the gradient, m, of the required line $L_{1}$ m = [ (($a^{2}$ - $b^{2}$)/c) - (c/3) ] $\div$ [ b - b/3 ] m = [ (3($a^{2}$ - $b^{2}$) - $c^{2}$)/3c ] $\div$ [ 2b/3 ] m = [ (3($a^{2}$ - $b^{2}$) - $c^{2}$)/3c ] $\times$ [ 3/(2b) ] m = (3($a^{2}$ - $b^{2}$) - $c^{2}$)/2bc Now to find the equation of the line, $L_{1}$ we simply plug in the required values into the following formula: y - $y_{1}$ = m(x - $x_{1}$), where y and x are variables, $y_{1}$ and $x_{1}$ are the y and x values, respectively, of a point on the line, and m is the gradient of the line. y - (c/3) = [(3($a^{2}$ - $b^{2}$) - $c^{2}$)/2bc] $\times$ [x - (b/3)] y = (((3($a^{2}$ - $b^{2}$) - $c^{2}$)x)/2bc) - ((3($a^{2}$ - $b^{2}$) - $c^{2}$)/6c) + c/3 y = (((3($a^{2}$ - $b^{2}$) - $c^{2}$)x)/2bc) - (3($a^{2}$ - $b^{2}$ - 3$c^{2}$))/6c Now substitute the third point, $P_{1}$, into the equation of the line $L_{1}$, to see if it satisfies it. (-$a^{2}$ + $b^{2}$ + $c^{2}$)/2c = - (3($a^{2}$ - $b^{2}$ - 3$c^{2}$))/6c Multiplying the right side of the equation by 3/3 gives: (-$a^{2}$ + $b^{2}$ + $c^{2}$)/2c = (-$a^{2}$ + $b^{2}$ + $c^{2}$)/2c As the equation is satisfied by all three points, they are found to be collinear.

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