Question

distances on the coordinate plane
find the distance between points a and b. state whether you worked with x or y coordinates.
then, show what addition or subtraction expression you used to find the distance.
1)
a(-3,2), b(-3,6)
6 - 2 = 4 units
2)
a(5,-6), b(5,3)
3 - (-6) = 9 units
3)
a(-8,5), b(-3,5)
-3 - (-8) = 5 units (note: original handwritten might have typo, corrected for sense)
4)
a(-6,-3), b(9,-3)
9 - (-6) = 15 units (note: original handwritten might have typo, corrected for sense)

Explanation:

Step1: Identify coordinates (1)

Points: $A(-7,2)$, $B(-7,6)$

Step2: Use y-coordinates, subtract

$\vert 6 - 2 \vert = 4$

Step3: Identify coordinates (2)

Points: $A(5,-6)$, $B(5,3)$

Step4: Use y-coordinates, subtract

$\vert 3 - (-6) \vert = \vert 3 + 6 \vert = 9$

Step5: Identify coordinates (3)

Points: $A(-5,5)$, $B(-3,5)$

Step6: Use x-coordinates, subtract

$\vert -3 - (-5) \vert = \vert -3 + 5 \vert = 2$

Step7: Identify coordinates (4)

Points: $A(-6,-3)$, $B(9,-5)$

Step8: Use distance formula

$$\sqrt{(9 - (-6))^2 + (-5 - (-3))^2} = \sqrt{(15)^2 + (-2)^2} = \sqrt{225 + 4} = \sqrt{229}$$

Answer:

1. Distance: 4 units; Used y-coordinates; Expression: $6-2$
2. Distance: 9 units; Used y-coordinates; Expression: $3-(-6)$
3. Distance: 2 units; Used x-coordinates; Expression: $-3-(-5)$
4. Distance: $\sqrt{229}$ units; Used both coordinates; Expression: $\sqrt{(9 - (-6))^2 + (-5 - (-3))^2}$