Question

What is the distance between the points (14, 29) and (14, 58) in the coordinate plane?

Answer 1

Answer: 29 units

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Explanation:
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$\textnormal {Formula : Distance }= \sqrt{(Y_2 - Y_1) ^2+(X_2 - X_1)^2}$

$\textnormal {Plug }(X_1 ,Y_1) = (14, 29) \textnormal { and } (X_2 , Y_2) = (14, 58) \textnormal { into the formula: }$

$\textnormal {Distance }= \sqrt{(58 -29) ^2+(14-14)^2} = \sqrt{29^2 + 0^2} = \sqrt{841} = 29 \textnormal { units}$

Answer 2

$Distance= \sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1})^2} = \sqrt{(58-29)^{2}+(14-14)^{2}} =29 But because x_{1}=x_{2}, and it is a vertical segment, it is enough to find abs(y_{2} -y_{1})=|58-29|=29$