Slopes of these:
(3,-2)(7,8)
(6,7)(8,-1)
(1,-5)(-2,9)
(-5,2)(-3,-7)
(6,3)(4,9)
(2,-5)(-3,9)
(-1,-7)(5,8)
(8,9)(-5,2)
(-4,-7)(-3,2)
(5,-9)(1,6)
This is only one question on my lesson by thee way.
The slope between (x1, y1) and (x2, y2) \(=\quad\dfrac{{\color{purple}y_2}-{\color{blue}y_1}}{{\color{purple}x_2}-{\color{blue}x_1}}\)
The slope between (3, -2) and (7, 8) \(=\quad\dfrac{{\color{purple}8}-{\color{blue}-2}}{{\color{purple}7}-{\color{blue}3}}\quad=\quad\dfrac{10}{4}\quad=\quad\dfrac52\)
The slope between (6, 7) and (8, -1) \(=\quad\dfrac{{\color{purple}-1}-{\color{blue}7}}{{\color{purple}8}-{\color{blue}6}}\quad=\quad\dfrac{-8}{2}\quad=\quad-4\)
The slope between (1, -5) and (-2, 9) \(=\quad\dfrac{{\color{purple}9}-{\color{blue}-5}}{{\color{purple}-2}-{\color{blue}1}}\quad=\quad\dfrac{14}{-3}\quad=\quad-\dfrac{14}{3}\)
The slope between (-5, 2) and (-3, -7) \(=\quad\dfrac{{\color{purple}-7}-{\color{blue}2}}{{\color{purple}-3}-{\color{blue}-5}}\quad=\quad\dfrac{-9}{2}\quad=\quad-\dfrac92\)
The slope between (6, 3) and (4, 9) \(=\quad\dfrac{{\color{purple}9}-{\color{blue}3}}{{\color{purple}4}-{\color{blue}6}}\quad=\quad\dfrac{6}{-2}\quad=\quad-3\)
Do you see how to do this? Can you do the rest?
The slope between (x1, y1) and (x2, y2) \(=\quad\dfrac{{\color{purple}y_2}-{\color{blue}y_1}}{{\color{purple}x_2}-{\color{blue}x_1}}\)
The slope between (3, -2) and (7, 8) \(=\quad\dfrac{{\color{purple}8}-{\color{blue}-2}}{{\color{purple}7}-{\color{blue}3}}\quad=\quad\dfrac{10}{4}\quad=\quad\dfrac52\)
The slope between (6, 7) and (8, -1) \(=\quad\dfrac{{\color{purple}-1}-{\color{blue}7}}{{\color{purple}8}-{\color{blue}6}}\quad=\quad\dfrac{-8}{2}\quad=\quad-4\)
The slope between (1, -5) and (-2, 9) \(=\quad\dfrac{{\color{purple}9}-{\color{blue}-5}}{{\color{purple}-2}-{\color{blue}1}}\quad=\quad\dfrac{14}{-3}\quad=\quad-\dfrac{14}{3}\)
The slope between (-5, 2) and (-3, -7) \(=\quad\dfrac{{\color{purple}-7}-{\color{blue}2}}{{\color{purple}-3}-{\color{blue}-5}}\quad=\quad\dfrac{-9}{2}\quad=\quad-\dfrac92\)
The slope between (6, 3) and (4, 9) \(=\quad\dfrac{{\color{purple}9}-{\color{blue}3}}{{\color{purple}4}-{\color{blue}6}}\quad=\quad\dfrac{6}{-2}\quad=\quad-3\)
Do you see how to do this? Can you do the rest?