Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - P.2 - Linear Models and Rates of Change - Exercises: 71


( b, ($a^{2}$-$b^{2}$)/c )

Work Step by Step

To find the point of intersection of the segments shown, we need to work out the equations for any two of them, and solve said equations simultaneously. This question involves a fair bit of intuition as the x and y axes are not marked. The base of the triangle is a horizontal line with both given y-coordinates equal to 0. This is therefore the x-axis. The perpendicular line $L_{1}$ passing through the point (b,c) therefore has the equation x = b, as it is parallel to the y-axis and has an infinite gradient. Equation of $L_{1}$: x = b Now we need to find the equation of either of the other two line segments. To do this we need the gradient of said segment and a point that it passes through. As a point has already been provided, we must only find the gradient in order to find the equation. However, neither of the two line segments available has two given points to calculate the gradient so we must determine it by taking the negative reciprocal of the gradient of a line perpendicular to either of them. I chose the line $L_{2}$ passing through the points (b,c) and (a,0). m($L_{2}$) = (c-0)/(b-a) m($L_{2}$) = c/(b-a), where m($L_{2}$) is the gradient of the line $L_{2}$. The negative reciprocal of m($L_{2}$) will give us the gradient of the line segment, $L_{3}$, passing through the point (-a,0) m($L_{3}$) = -(b-a)/c m($L_{3}$) = (a-b)/c, where m($L_{3}$) is the gradient of the line segment $L_{3}$ Now to find the equation of the line segment $L_{3}$ 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 x and y values of a point on the line and m is the gradient of the line. Given the point (-a,0) $y_{1}$ = 0 $x_{1}$ = -a m = m($L_{3}$) = (a-b)/c The equation of the line $L_{3}$ is then given by: y - 0 = ((a-b)/c)(x+a) y = [(a-b)(x+a)]/c Substituting Equation $L_{1}$ into $L_{3}$ we get: y = [(a-b)(a+b)]/c y = ($a^{2}$ - $b^{2}$)/c Therefore the coordinates of the point of intersection of the line segments are: ( b , ($a^{2}$ - $b^{2}$)/c )
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