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saul85 [17]
3 years ago
5

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

Mathematics
2 answers:
Snowcat [4.5K]3 years ago
7 0
Answer: 29 units

∞∞∞∞∞∞∞∞∞
<em>Explanation:</em>
∞∞∞∞∞∞∞∞∞

\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}


ss7ja [257]3 years ago
6 0
Distance= \sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1})^2} = \sqrt{(58-29)^{2}+(14-14)^{2}} =29&#10;&#10;But because x_{1}=x_{2}, &#10;&#10; and it is a vertical segment, &#10;&#10;it is enough to find abs(y_{2} -y_{1})=|58-29|=29
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Solution :-

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