The two opposite vertices of a square are $(- 1, 2)$ and $(3, 2)$ . Find the coordinates of the other two vertices.

Question

The two opposite vertices of a square are (−1,2) and (3,2). Find the coordinates of the other two vertices.

Filo · Accepted Answer

Let ABCD is a square where two opposite vertices are A(−1,2)andC(3,2).Let B(x,y) and D(x1​.y1​) ids the other two vertices.In Square ABCDAB=BC=CD=DAHence AB=BC⇒(x+1)2+(y−2)2​=(3−x)2+(2−y)2​   [by distance formula]Squaring both sides⇒(x+1)2+(y−2)2=(3−x)2+(2−y)2⇒x2+2x+1+y2+4−4y=9+x2−6x+4+y2−4y⇒2x+5=13−6x⇒2x+6x=13−5⇒8x=8⇒x=1In △ABC,∠B=90∘  [all angles of square are 90∘]Then according to the Pythagorean theoremAB2+BC2=AC2As AB=BC∴2AB2=AC2⇒2(x+1)2+(y−2)2​)2=((3−(−1))2+(2−2)2​)2⇒2((x+1)2+(y−2)2)=(3+1)2+(2−2)2⇒2(x2+2x+1+y2+4−4y)=(4)2Put the value of x=1⇒2(11+2×1+1+y2+4−4y)=16⇒2(y2−4y+8)=16⇒2y2−8y+16=16⇒2y2−8y=0⇒2y(y−4)=0⇒(y−4)=0Hence y=0 or 4.As diagonals of a square are equal in length and bisect each other at 90∘Let P is the midpoint of AC ∴CO-ordinates of P=(23−1​,22+2​)=(1,2)P is also the midpoint of BDthen co-ordinates of mid-point of BD=co-ordinates of P⇒(x1​,y1​)=1,2∴x1​=1,y1​=2Then other two vertices of square ABCD are (1,0) or (1,4) and (1,2).

Question Text	The two opposite vertices of a square are $(- 1, 2)$ and $(3, 2)$ . Find the coordinates of the other two vertices.
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The two opposite vertices of a square are $(- 1, 2)$ and $(3, 2)$ . Find the coordinates of the other two vertices.

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