Square

Points A[-9,7] and B[-4,-5] are adjacent vertices of the square ABCD. Calculate the area of the square ABCD.

Result

S =  169

Solution:

a2=Δx2+Δy2  x0=−9 y0=7 x1=−4 y1=−5  a=(x0−x1)2+(y0−y1)2=((−9)−(−4))2+(7−(−5))2=13  S=a2=132=169a^2=\Delta x^2 + \Delta y^2 \ \\ \ \\ x_{ 0 }=-9 \ \\ y_{ 0 }=7 \ \\ x_{ 1 }=-4 \ \\ y_{ 1 }=-5 \ \\ \ \\ a=\sqrt{ (x_{ 0 }-x_{ 1 })^2+(y_{ 0 }-y_{ 1 })^2 }=\sqrt{ ((-9)-(-4))^2+(7-(-5))^2 }=13 \ \\ \ \\ S=a^2=13^2=169

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