If ABCD is a kite, then by definition it has two pairs of congruent sides.
Let point C has coordinates (a,b).
Then
- [tex]AB=\sqrt{(-1-0)^2+(5-6)^2}=\sqrt{1+1}=\sqrt{2};[/tex]
- [tex]BC=\sqrt{(a-0)^2+(b-6)^2};[/tex]
- [tex]CD=\sqrt{(a-0)^2+(b-2)^2};[/tex]
- [tex]AD=\sqrt{(-1-0)^2+(5-2)^2}=\sqrt{1+9}=\sqrt{10}.[/tex]
Solve the system of equations:
[tex]\left\{\begin{array}{l}\sqrt{(a-0)^2+(b-6)^2}=\sqrt{2}\\\sqrt{(a-0)^2+(b-2)^2}=\sqrt{10}\end{array}\right.[/tex]
Square these two equations and then subtract:
[tex](b-6)^2-(b-2)^2=2-10,\\ \\b^2-12b+36-b^2+4b-4=-8,\\ \\-8b=-8-32,\\ \\-8b=-40,\\ \\b=5.[/tex]
Substitute b=5 into the first equation:
[tex]\sqrt{a^2+(5-6)^2}=\sqrt{2},\\ \\a^2=2-1,\\ \\a^2=1,\\ \\a=1 \text{ or } a=-1.[/tex]
You get two points (1,5) and (-1,5). Point (-1,5) coincides with point A, so C(1,5).
Answer: correct choice is B.
