6. slopes of perpendicular lines are negative reciprocals of each other. Therefore if CD is [tex] \frac{5}{2} [/tex] , the slope of AB will be - [tex] \frac{2}{5} [/tex]
7. Note that the y intercept of the line is (0,8). Also note that another point on the line is (3,4). Now use these two points to find the slope [tex] \frac{rise}{run} [/tex] Moving from point (0,8) to (3,4), we go down 4 and to the right 3. Therefore, the slope is - [tex] \frac{4}{3} [/tex]. At this point, we can just substitute into slope intercept form of a line: y = mx + b or [tex]y = - \frac{4}{3} x + 8 [/tex]
8. First find the slope of the line containing the points (-2,-3) and (2,5): The slope is [tex] \frac{5 - (-3))}{2- (-2))} = \frac{8}{4} = 2[/tex] Now substitute that slope and the point (-4,5) into point slope form of the a line. You get:
y - 5 = 2(x + 4)
