Find the 3

Find the distance and mid-point between A(1,2) and B(5,5).

Result

d =  5
x =  3
y =  3.5

Solution:

x0=1 y0=2  x1=5 y1=5  d=(x1−x0)2+(y1−y0)2=(5−1)2+(5−2)2=5x_{0}=1 \ \\ y_{0}=2 \ \\ \ \\ x_{1}=5 \ \\ y_{1}=5 \ \\ \ \\ d=\sqrt{ (x_{1}-x_{0})^2+(y_{1}-y_{0})^2 }=\sqrt{ (5-1)^2+(5-2)^2 }=5
x=(x0+x1)/2=(1+5)/2=3x=(x_{0}+x_{1})/2=(1+5)/2=3
y=(y0+y1)/2=(2+5)/2=72=3.5y=(y_{0}+y_{1})/2=(2+5)/2=\dfrac{ 7 }{ 2 }=3.5

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