# 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:

$a^2=\Delta x^2 + \Delta y^2 \ \\ \ \\ x_{ 0 }=-9 \ \\ y_{ 0 }=7 \ \\ x_{ 1 }=-4 \ \\ y_{ 1 }=-5 \ \\ \ \\ a=\sqrt{ (x_{ 0 }-x_{ 1 })^2+(y_{ 0 }-y_{ 1 })^2 }=\sqrt{ ((-9)-(-4))^2+(7-(-5))^2 }=13 \ \\ \ \\ S=a^2=13^2=169$

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