Slope worksheet

1. Slope (m): 1

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{8-3}{7-2}$$=\frac{5}{5}$$=1$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=3-1\times 2$$=1$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = x + 1}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(1)}$$=45$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{5^2+5^2}$$=\sqrt{50}$$=7.071068$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((2 + 7) ÷ 2, (3 + 8) ÷ 2)}$$=\text{(4.5, 5.5)}$

2. Slope (m): 2.75

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{13-2}{6-2}$$=\frac{11}{4}$$=2.75$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=2-2.75\times 2$$=\mathrm{-3.5}$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = 2.75x − 3.5}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(2.75)}$$=70.016893$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{4^2+{11}^2}$$=\sqrt{137}$$=11.7047$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((2 + 6) ÷ 2, (2 + 13) ÷ 2)}$$=\text{(4, 7.5)}$

3. Slope (m): 3

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{9-3}{5-3}$$=\frac{6}{2}$$=3$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=3-3\times 3$$=\mathrm{-6}$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = 3x − 6}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(3)}$$=71.565051$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{2^2+6^2}$$=\sqrt{40}$$=6.324555$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((3 + 5) ÷ 2, (3 + 9) ÷ 2)}$$=\text{(4, 6)}$

4. Slope (m): 1.43

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{13-3}{9-2}$$=\frac{10}{7}$$=1.428571$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=3-1.428571\times 2$$=0.142857$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = 1.428571x + 0.142857}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(1.428571)}$$=55.00798$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{7^2+{10}^2}$$=\sqrt{149}$$=12.206556$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((2 + 9) ÷ 2, (3 + 13) ÷ 2)}$$=\text{(5.5, 8)}$

5. Slope (m): 1.5

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{9-3}{7-3}$$=\frac{6}{4}$$=1.5$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=3-1.5\times 3$$=\mathrm{-1.5}$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = 1.5x − 1.5}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(1.5)}$$=56.309932$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{4^2+6^2}$$=\sqrt{52}$$=7.211103$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((3 + 7) ÷ 2, (3 + 9) ÷ 2)}$$=\text{(5, 6)}$

6. Slope (m): 1.75

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{11-4}{6-2}$$=\frac{7}{4}$$=1.75$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=4-1.75\times 2$$=0.5$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = 1.75x + 0.5}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(1.75)}$$=60.255119$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{4^2+7^2}$$=\sqrt{65}$$=8.062258$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((2 + 6) ÷ 2, (4 + 11) ÷ 2)}$$=\text{(4, 7.5)}$

7. Slope (m): 1.8

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{12-3}{6-1}$$=\frac{9}{5}$$=1.8$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=3-1.8\times 1$$=1.2$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = 1.8x + 1.2}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(1.8)}$$=60.945396$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{5^2+9^2}$$=\sqrt{106}$$=10.29563$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((1 + 6) ÷ 2, (3 + 12) ÷ 2)}$$=\text{(3.5, 7.5)}$

8. Slope (m): 1.4

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{10-3}{7-2}$$=\frac{7}{5}$$=1.4$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=3-1.4\times 2$$=0.2$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = 1.4x + 0.2}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(1.4)}$$=54.462322$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{5^2+7^2}$$=\sqrt{74}$$=8.602325$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((2 + 7) ÷ 2, (3 + 10) ÷ 2)}$$=\text{(4.5, 6.5)}$

9. Slope (m): 4

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{11-3}{5-3}$$=\frac{8}{2}$$=4$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=3-4\times 3$$=\mathrm{-9}$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = 4x − 9}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(4)}$$=75.963757$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{2^2+8^2}$$=\sqrt{68}$$=8.246211$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((3 + 5) ÷ 2, (3 + 11) ÷ 2)}$$=\text{(4, 7)}$

10. Slope (m): 2

1. Slope (m) is rise over run: how far the line goes up (change in y) divided by how far it goes across (change in x)

   $m$$=\frac{11-3}{6-2}$$=\frac{8}{4}$$=2$

2. Find where the line crosses the y-axis (b): start from the first point and undo the slope back to x = 0

   $b$$=3-2\times 2$$=\mathrm{-1}$

3. Put the slope m and the crossing point b into the line's equation, y = mx + b

   $\text{y = 2x − 1}$

4. Find the angle the line makes with the x-axis, using the arctangent (the angle whose slope this is)

   $\theta$$=\text{arctan(2)}$$=63.434949$

5. Find the distance between the points. The run and rise form the two short sides of a right triangle, so the Pythagorean theorem gives the long side

   $d$$=\sqrt{4^2+8^2}$$=\sqrt{80}$$=8.944272$

6. Find the midpoint, halfway between: average the two x values and the two y values

   $M$$=\text{((2 + 6) ÷ 2, (3 + 11) ÷ 2)}$$=\text{(4, 7)}$