QUESTION IMAGE
Question
drag each set of coordinates to the correct location on the table. calculate the distance between the pairs of coordinates, and classify them according to the distance between them. (3, 4) and (2, 1) (3, 7) and (5, 2) (5, -2) and (3, 3) (-2, 3) and (1, -2) (-4, -2) and (-3, 1) (4, -1) and (-1, 1) √10 units √29 units
Step1: Recall distance formula
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
Step2: Calculate distance for $(3,4)$ and $(2,1)$
$d=\sqrt{(2 - 3)^2+(1 - 4)^2}=\sqrt{(-1)^2+(-3)^2}=\sqrt{1 + 9}=\sqrt{10}$
Step3: Calculate distance for $(3,7)$ and $(5,2)$
$d=\sqrt{(5 - 3)^2+(2 - 7)^2}=\sqrt{2^2+(-5)^2}=\sqrt{4 + 25}=\sqrt{29}$
Step4: Calculate distance for $(5,-2)$ and $(3,3)$
$d=\sqrt{(3 - 5)^2+(3+ 2)^2}=\sqrt{(-2)^2+5^2}=\sqrt{4 + 25}=\sqrt{29}$
Step5: Calculate distance for $(-2,3)$ and $(1,-2)$
$d=\sqrt{(1 + 2)^2+(-2 - 3)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9 + 25}=\sqrt{34}$
Step6: Calculate distance for $(-4,-2)$ and $(-3,1)$
$d=\sqrt{(-3 + 4)^2+(1 + 2)^2}=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}$
Step7: Calculate distance for $(4,-1)$ and $(-1,1)$
$d=\sqrt{(-1 - 4)^2+(1 + 1)^2}=\sqrt{(-5)^2+2^2}=\sqrt{25 + 4}=\sqrt{29}$
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$\sqrt{10}$ units: $(3,4)$ and $(2,1)$, $(-4,-2)$ and $(-3,1)$
$\sqrt{29}$ units: $(3,7)$ and $(5,2)$, $(5,-2)$ and $(3,3)$, $(4,-1)$ and $(-1,1)$