Explanation Procedure used: 1. Solve the given equation for y . 2. Compute the slope m from the above obtained equation. 3. The slope of the parallel line and the given line are equal. 4. Obtain the point-slope form of the straight line by substituting the value of m x 1 and y 1 in y − y 1 = m ( x − x 1 ) . 5. Express the equation in slope intercept form y = m x + b or in the standard form A x + B y = C . Calculation: From the given information, the values of x 1 and y 1 are − 4 and − 2 , respectively. Compute the equation of the parallel line as follows. Solve the given equation x + 3 y = 6 for y as follows. 3 y = − x + 6 3 y 3 = − x + 6 3 y = − x 3 + 6 3 y = − x 3 + 2 Compare the above equation with the slope –intercept form y = m x + b and obtain the value of m . y = m ↓ x + b y = − x 3 + 2 From the above equation, the value of m is − 1 3 . Now substitute x 1 = − 4 , y 1 = − 2 and m = − 1 3 , in y − y 1 = m ( x − x 1 ) . y − ( − 2 ) = − 1 3 ( x − ( − 4 ) ) 3 ( y + 2 ) = − 1 ( x + 4 ) 3 y + 6 = − x − 4

MML F/COLLEGE MAT F/TRADES - ACCESS CODE

10th Edition

ISBN: 9781323845967

Author: Hobbs

Publisher: PEARSON

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Chapter 9, Problem 141RE

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The standard form of the equation of the line that is parallel to the line x+3y=6 and passes through the point (−4,−2).

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Chapter 9, Problem 140RE

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MML F/COLLEGE MAT F/TRADES - ACCESS CODE

Ch. 9.1 - Find y if x = 4 in the equation 3x + 2y = 6. Ch. 9.1 - Locate the points in Fig. 9–12 by giving the...Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E

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The standard form of the equation of the line that is parallel to the line x + 3 y = 6 and passes through the point ( − 4 , − 2 ) .

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The standard form of the equation of the line that is parallel to the line x+3y=6 and passes through the point (−4,−2).

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