Prove that the points A(-5,4), B(-1.-2) and C(5,2) are the vertices of an isosceles right angled triangle. Also, find the coordinates of D, so that ABCD is a square.

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Prove that the points A(-5,4), B(-1.-2) and C(5,2) are the vertices of an isosceles right angled triangle. Also, find the coordinates of D, so that ABCD is a square.

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<div class = 'in102'>We must prove that A(-5,4 ), B(-1,2 ) and C(5,2 ) are the vertices of isosceles right triangles. use the distance formula AB = $\sqrt{}(-5+1)^2+(4+2)^2{}=\sqrt{}16+36{}=2\sqrt{}13{}$BC = $\sqrt{}(-1-5)^2+(-2-2)^2{}=\sqrt{}36+16{}=2\sqrt{}13{}$CA=$\sqrt{}(5+5)^2+(2-4)^2{}=\sqrt{}100+4{}=2\sqrt{}26{}$here, obviously AB2 + BC2 = 52 + 52 = 104 = CA2. From the Pythagoras theorem, we know that any triangle will be a right triangle, and the triangle on the side follows the above conditions. abc is a right triangle Let D(x, y ) make ABCD a square.We know that squares are also parallelograms. So, AC's diagonal midpoint = diagonal BD's midpoint. {}(5 - 5)/2, (4 + 2)/2{} = {}(x - 1)/2, (y - 2)/2{} or, ( 0,3)={}(x-1)/2,(y-2)/2{} ( x-1)/2 = 0 = & gt;x = 1. and ( y-2)/2=3=>y=8Therefore, D = ( 1, 8).

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Prove that the points A(-5,4), B(-1.-2) and C(5,2) are the vertices of an isosceles right angled triangle. Also, find the coordinates of D, so that ABCD is a square.

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