256 Chapter 3 Linear Equations in Two Variables 4b. We first find the slope of y = 5x - 6. Since the equation is in slope-intercept form, the slope of the line is m = 5. The lines are perpendicular, so the slope of the unknown line is the negative reciprocal of 5. Hence, the slope of the unknown line is m=- 1 5 . Now we find the equation of the line with m=- 1 5 that passes through (-2, 4). y - y1 = m(x - x1) y - (4) = - 1 5 x - (-2) Let m=- 1 5 , (x1, y1) = (-2, 4). y - 4=- 1 5 (x + 2) Simplify. y - 4 = - 1 5 x - 2 5 Apply the distributive property. 5(y - 4) = 5a- 1 5 x - 2 5 b Multiply each side by 5. 5y - 20=-x - 2 Apply the distributive property. 5y - 20 + x=-x - 2 + x Add x to each side. x + 5y - 20=-2 Simplify. x + 5y - 20 + 20=-2 + 20 Add 20 to each side. x + 5y = 18 Simplify. So, the equation of the line through (-2, 4) and perpendicular to y = 5x - 6 is x + 5y = 18. 4c. We first find the slope of x = 5. The equation x = 5 represents a vertical line and, therefore, has undefined slope. Since the lines are parallel, the slope of the unknown line is also undefined. The unknown line is a vertical line that passes through the point (2, -3). So, its equation is x = -2. 4d. From part c, the slope of x = 5 is undefined. Since the lines are perpendicular, the slope of the unknown line is the negative reciprocal of the slope of the given line. Because the slope of the given line is undefined, we can’t find its negative reciprocal. We do know, however, that a line perpendicular to a vertical line is a horizontal line with slope m = 0. So, the equation of the horizontal line through (4, -7) is y = -7. Student Check 4 Write the equation of the line that goes through (2, -3) that is a. parallel to y = 1 2 x +6 b. perpendicular to y = 1 2 x + 6 c. parallel to x = -7 d. perpendicular to x = -7 Applications There are two types of application problems that we will examine. One type involves knowing an initial value (or beginning value, that is, the y-value that corresponds to x = 0) and information about how that value changes. The other type involves knowing two data points that describe a particular situation. Objective 5 ▶ Solve application problems.
hendricks_beginning_algebra_1e_ch1_3
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