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According to Pythagoras theorem, distance between points (-3, 8) and (8, -5) is

 Jan 20, 2020
 #1
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According to Pythagoras theorem, distance between points (-3, 8) and (8, -5) is

 

\(\text{Let $P_1(x_1,~y_1)=(-3,~8)$} \\ \text{Let $P_2(x_2,~y_2)=(8,~-5)$}\)

 

Formula: \(\text{distance} = \sqrt{ \left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\text{distance}} &=& \mathbf{\sqrt{ \left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}} \\\\ \text{distance} &=& \sqrt{ \Big(8-(-3)\Big)^2+\left(-5-8\right)^2} \\\\ \text{distance} &=& \sqrt{ 11^2+\left(-13\right)^2} \\\\ \text{distance} &=& \sqrt{ 121+169} \\\\ \text{distance} &=& \sqrt{ 290} \\\\ \text{distance} &=& 17.0293863659 \\ \hline \end{array} \)

 

The distance is 17.0

 

laugh

 Jan 20, 2020

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