257 2.5 Equations of Lines and Linear Models Standard form EXAMPLE 6 Finding Equations of Parallel and Perpendicular Lines Write an equation in both slope-intercept and standard form of the line that passes through the point 13, 52 and satisfies the given condition. (a) parallel to the line 2x + 5y = 4 (b) perpendicular to the line 2x + 5y = 4 SOLUTION (a) We know that the point 13, 52 is on the line, so we need only find the slope to use the point-slope form. We find the slope by writing the equation of the given line in slope-intercept form—that is, we solve for y. 2x + 5y = 4 5y = -2x + 4 Subtract 2x. y = - 2 5 x + 4 5 Divide by 5. The slope is - 2 5 . Because the lines are parallel, - 2 5 is also the slope of the line whose equation is to be found. Now substitute this slope and the given point 13, 52 in the point-slope form. y - y1 = m1x - x12 Point-slope form y - 5 = - 2 5 1x - 32 m= - 2 5 , x1 = 3, y1 = 5 y - 5 = - 2 5 x + 6 5 Distributive property y = - 2 5 x + 31 5 Add 5 = 25 5 . 5y = -2x + 31 Multiply by 5. 2x + 5y = 31 Add 2x. (b) There is no need to find the slope again—in part (a) we found that the slope of the line 2x + 5y = 4 is - 2 5 . The slope of any line perpendicular to it is 5 2 . y - y1 = m1x - x12 Point-slope form y - 5 = 5 2 1x - 32 m= 5 2 , x1 = 3, y1 = 5 y - 5 = 5 2 x - 15 2 Distributive property y = 5 2 x - 5 2 Add 5 = 10 2 . 2y = 5x - 5 Multiply by 2. -5x + 2y = -5 Subtract 5x. 5x - 2y = 5 Multiply by -1 so that A70. S Now Try Exercises 51 and 53. Slope-intercept form Slope-intercept form Standard form
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