164 Chapter 2 Linear Equations in Two Variables and Functions Now find an equation of the line passing through ( ) having a slope y y1 m1x x12 1x1, y11 2. y m 4 2, 32 132 43x 122 4 4, 1 2x y 7. y 3 41x 22 y 3 4x 8 y 4x 11 Finding an Equation of a Line Perpendicular to Another Line Example 8 Find an equation of the line passing through the point (4, 3) and perpendicular to the line Write the answer in slope-intercept form. Solution: The slope of the given line can be found from its slope-intercept form. Solve for y. 2x 3y 3. 2x 3y 3 3y 2x 3 2x 3 2 3 3y 3 3 3 y . The slope is 2 3 x 1 The slope of a line perpendicular to this line must be the opposite of the m 32 m 32 23; . reciprocal of hence, Using and the known point (4, 3), we can apply the point-slope formula to find an equation of the line. Apply the point-slope formula. m 32 Substitute and (4, 3) for (x1, y1). Clear parentheses. Solve for y. y y1 m1x x12 y 3 y 3 y 3 2 3 2 3 2 1x 42 x 6 x 3 Skill Practice 10. Find an equation of the line passing through the point ( 1,6 ) and perpendicular to the line x 2y 8.Write the answer in slope-intercept form. of Apply the point-slope formula. Substitute and for Clear parentheses. Write the answer in slope-intercept form. Skill Practice 9. Find an equation of a line containing ( ) and parallel to Write the answer in slope-intercept form. We can verify the answer to Example 7 by graphing both equations.We see that the line defined by y 4x 11 passes through the point ( 2, 3 ) and is parallel to the line y 4x 8. See Figure 2-24. 4. 2, 3 6 4 42 12 Answers 9. 10. y 2x 8 y 2x 9 x y 12 10 8 2 121086 2 4 6 8 10 2 4 6 8 10 12 y 4x 11 (2, 3) y 4x 8 Figure 2-24
miller_intermediate_algebra_4e_ch1_3
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