Let's proceed step by step to show that the quadrilateral KITE, with vertices [tex]\(K (0,-2)\)[/tex], [tex]\(I (1,2)\)[/tex], [tex]\(T (7,5)\)[/tex], and [tex]\(E (4,-1)\)[/tex], is a kite.
1. Calculate [tex]\(KI\)[/tex]:
Using the distance formula:
[tex]\[
KI = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
where [tex]\(K = (0, -2)\)[/tex] and [tex]\(I = (1, 2)\)[/tex]:
[tex]\[
KI = \sqrt{(1 - 0)^2 + (2 - (-2))^2} = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}
\][/tex]
2. Calculate [tex]\(KE\)[/tex]:
Using the distance formula for [tex]\(K = (0, -2)\)[/tex] and [tex]\(E = (4, -1)\)[/tex]:
[tex]\[
KE = \sqrt{(4 - 0)^2 + (-1 - (-2))^2} = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17}
\][/tex]
3. Calculate [tex]\(IT\)[/tex]:
For [tex]\(I = (1, 2)\)[/tex] and [tex]\(T = (7, 5)\)[/tex]:
[tex]\[
IT = \sqrt{(7 - 1)^2 + (5 - 2)^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}
\][/tex]
4. Calculate [tex]\(TE\)[/tex]:
For [tex]\(T = (7, 5)\)[/tex] and [tex]\(E = (4, -1)\)[/tex]:
[tex]\[
TE = \sqrt{(4 - 7)^2 + (-1 - 5)^2} = \sqrt{(-3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}
\][/tex]
Given the calculated distances:
- [tex]\( KI = \sqrt{17} \)[/tex]
- [tex]\( KE = \sqrt{17} \)[/tex]
- [tex]\( IT = 3\sqrt{5} \)[/tex]
- [tex]\( TE = 3\sqrt{5} \)[/tex]
Therefore, [tex]\(KI\)[/tex] and [tex]\(KE\)[/tex] are equal, and so are [tex]\(IT\)[/tex] and [tex]\(TE\)[/tex]. Since two pairs of adjacent sides are of equal length, quadrilateral KITE is a kite.
So, filling in the drop-down menus:
- [tex]\(KE = \sqrt{17}\)[/tex]
- [tex]\(IT = 3\sqrt{5}\)[/tex]
- [tex]\(TE = 3\sqrt{5}\)[/tex]
- Therefore, KITE is a kite because it has two pairs of adjacent sides that are equal in length.
