Question

20%280%2C%202%5C%5C2...%282%2C%203%29%20%26%20%28-3%2C%200%29%5C%5C3...%28-4%2C%204%29%20%26%20%284%2C%205%29%5C%5C4...%285%2C%203%29%20%26%20%28-1%2C%205%29%5C%5C5...%28-3%2C%203%29%20%26%20%280.%20-1%29" id="TexFormula1" title="Write\;equations\;for\;the\;following...\\\\1...(-2, 3) & (0, 2\\2...(2, 3) & (-3, 0)\\3...(-4, 4) & (4, 5)\\4...(5, 3) & (-1, 5)\\5...(-3, 3) & (0. -1)" alt="Write\;equations\;for\;the\;following...\\\\1...(-2, 3) & (0, 2\\2...(2, 3) & (-3, 0)\\3...(-4, 4) & (4, 5)\\4...(5, 3) & (-1, 5)\\5...(-3, 3) & (0. -1)" align="absmiddle" class="latex-formula">

Answer 1

Answer:  see below

Step-by-step explanation:

Step 1: Find the slope (m) using the formula:    $m=\dfrac{y_2-y_1}{x_2-x_1}$

Step 2: Plug in the slope (m) and one of the points (x₁, y₁) into the Point-Slope formula.      y - y₁ = m(x - x₁)

1. (-2, 3) and (0, 2)

Step 1:  $m=\dfrac{3-2}{-2-0}\quad =\dfrac{1}{-2}\quad =-\dfrac{1}{2}$

Step 2:  $y-2=-\dfrac{1}{2}(x-0)$

             $y-2=-\dfrac{1}{2}x$

                 $\large\boxed{y=-\dfrac{1}{2}x+2}$

2. (2, 3) and (-3, 0)

Step 1:  $m=\dfrac{3-0}{2-(-3)}\quad =\dfrac{3-0}{2+3}\quad =\dfrac{3}{5}$

Step 2:  $y-0=\dfrac{3}{5}(x-(-3))$

                   $y=\dfrac{3}{5}(x+3)$

                 $\large\boxed{y=\dfrac{3}{5}x+\dfrac{9}{5}}$

3. (-4, 4) and (4, 5)

Step 1:  $m=\dfrac{4-5}{-4-4}\quad =\dfrac{-1}{-8}\quad =\dfrac{1}{8}$

Step 2:  $y-4=\dfrac{1}{8}(x-(-4))$

             $y-4=\dfrac{1}{8}(x+4)$

             $y-4=\dfrac{1}{8}x+\dfrac{1}{2}$

                 $\large\boxed{y=\dfrac{1}{8}x+\dfrac{9}{2}}$

4. (5, 3) and (-1, 5)

Step 1:  $m=\dfrac{3-5}{5-(-1)}\quad =\dfrac{3-5}{5+1}\quad =\dfrac{-2}{6}\quad =-\dfrac{1}{3}$

Step 2:  $y-3=-\dfrac{1}{3}(x-5)$

             $y-3=-\dfrac{1}{3}x+\dfrac{5}{3}$

                 $\large\boxed{y=-\dfrac{1}{3}x+\dfrac{14}{3}}$

5. (-3, 3) and (0, -1)

Step 1:  $m=\dfrac{3-(-1)}{-3-0}\quad =\dfrac{3+1}{-3-0}\quad =\dfrac{4}{-3}\quad =-\dfrac{4}{3}$

Step 2:  $y-3=-\dfrac{4}{3}(x-(-3))$

             $y-3=-\dfrac{4}{3}(x+3)$

             $y-3=-\dfrac{4}{3}x-4$

                 $\large\boxed{y=-\dfrac{4}{3}x-1}$

Answer 2

You can comment below, are these supposed to be simply equations or piecewise functions?