Two opposite vertices of a square are(-1,2)and(3,2) find the coordinate of other two vertices.

Two opposite vertices of a square are(-1,2)and(3,2) find the coordinat

1.	Two opposite vertices of a square are(-1,2)and(3,2) find the coordinate of other two vertices.
Answer» Let ABCD be a square and B (x, y) be the unknown vertex.AB = BC{tex} \\Rightarrow {/tex} AB2 = BC2\xa0{tex} \\Rightarrow {/tex}\xa0(x + 1)2 + (y - 2)2 = (x - 3)2 + (y - 2)2{tex} \\Rightarrow {/tex} x2 + 1 + 2x + y2 + 4 - 4x = x2 - 6x + 9 + y2 + 4 - 4x{tex} \\Rightarrow {/tex}\xa02x + 1 = - 6x + 9{tex} \\Rightarrow {/tex}\xa08x = 8{tex} \\Rightarrow {/tex}\xa0x = 1 ........ (i)In {tex}\\triangle{/tex}ABC, AB2 + BC2 = AC2{tex} \\Rightarrow {/tex}\xa0(x + 1)2 + (y - 2)2 + (x - 3)2 + (y - 2)2 = (3 + 1)2 + (2 - 2)2{tex} \\Rightarrow {/tex}x2 + 1 + 2x + y2 - 4x + 4 + x2 + 9 - 6x + y2 + 4 - 2y = 16 + 0{tex} \\Rightarrow {/tex}\xa02x2 + 2y2 + 2x - 4y - 6x - 4y + 1 + 4 + 9 + 4 = 16{tex} \\Rightarrow {/tex}\xa02x2 + 2y2 - 4x - 8y + 2 = 0{tex} \\Rightarrow {/tex}\xa0x2 + y2 - 2x - 4y + 1 = 0 ..... (ii)Putting the value of x in eq. (ii),1 + y2 - 2 - 4y + 1 = 0\xa0{tex} \\Rightarrow {/tex}\xa0y2 - 4y = 0\xa0{tex} \\Rightarrow {/tex}\xa0y(y - 4) = 0{tex} \\Rightarrow {/tex}\xa0y = 0 or 4Hence the other vertices are (1, 0) and (1, 4).