3 12 8. To determine whether the pair of lines is parallel ..., Schemes and Mind Maps of Calculus

Download 3 12 8. To determine whether the pair of lines is parallel ... and more Schemes and Mind Maps Calculus in PDF only on Docsity! 1.2 (page 28) 3 12 8. To determine whether the pair of lines is parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their slopes, and, if necessary, compare their y-intercepts. : 4 : 3 3 12 4 3 4 L x y M x y y x y x y x + = − + = − = − − = − − = − − slope: m = −1; y-intercept: (0, −4) slope: m = −1; y-intercept: (0, −4 ) Since both the slopes the y-intercepts of the two lines are the same, the lines are coincident. 10. To determine whether the pair of lines is parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their slopes, and, if necessary, compare their y-intercepts. : 4 2 7 : 2 2 2 4 7 2 2 7 2 2 L x y M x y y x y x y x − = − − + = − − = − − = − = + slope: m = 2; y-intercept: 70, 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ slope: m = 2; y-intercept: (0, −2) The slopes of the two lines are the same, but the y-intercepts are different, so the lines are parallel. 12. To determine whether the pair of lines is parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their slopes, and, if necessary, compare their y-intercepts. : 4 3 2 3 4 2 4 2 3 3 L x y y x y x + = = − + = − + : 2 1 2 1 2 1 M x y y x y x − = − − = − − = + slope: m = − 4 3 slope: m = 2 Since the slopes of the two lines are different, the lines intersect. 20. To find the point of intersection of two lines, first put the lines in slope-intercept form. : 4 2 8 2 4 L x y y x − = = − : 6 3 0 2 M x y y x + = = − Since the point of intersection, (x0, y0), must be on both L and M, we set the two equations equal to each other and solve for x0. Then we substitute the value of x0 into the equation of one of the lines to find y0. 0 0 0 0 2 4 2 4 4 1 x x x x − = − = = 0 0 2(1) 2 y y = − = − The point of intersection is (1, −2). 25. L is the vertical line on which the x-value is always 4. M is the horizontal line on which y-value is always −2. The point of intersection is (4, −2). L M 1 4 x 2 1 (4, 2) 2 5 4 y 27. L is parallel to y = 2x, so the slope of L is m = 2. We are given the point (3, 3) on line L. Use the point-slope form of the line. ( ) ( ) 1 1 3 2 3 3 2 6 y y m x x y x y x − = − − = − − = − slope-intercept form: 2 3y x= − general form: 2 3x y− = 30. We want a line parallel to y = −3x. So our line will have slope m = −3. It must also contain the point (−1, 2). Use the point slope form of the equation of a line: ( ) ( ) 1 1 2 3 1 2 3 3 y y m x x y x y x − = − − = − + − = − − slope-intercept form: 3 1y x= − − general form: 3 1x y+ = − 36. To find the equation of the line parallel to the line containing the points (−4, 5) and (2, −1), first find the slope of the line containing the two points. ( ) ( ) 2 1 2 1 5 1 6 1 4 2 6 y ym x x − −− = = = = − − − − − The slope of a line parallel to the line containing these points is also –1. Use the slope and the point (−2, −5) to write the point-slope form of the parallel line. ( ) ( ) 1 1 5 1 2 5 2 y y m x x y x y x − = − + = − + + = − − Solve for y to get the slope-intercept form: 7y x= − − Rearrange terms to obtain the general form of the equation: x + y = –7