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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Step-by-step explanation:
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`\textnormal {Formula : Distance }= √((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 }= √((58 -29) ^2+(14-14)^2) = √(29^2 + 0^2) = √(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`