Answer :
To show that quadrilateral KITE with vertices [tex]\(K (0, -2), I (1, 2), T (7, 5)\)[/tex], and [tex]\(E (4, -1)\)[/tex] is a kite, we need to verify specific properties of the distances between its vertices.
Step-by-Step Solution:
1. Calculate the distance [tex]\(K I\)[/tex]:
The distance formula is:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
For vertices [tex]\(K (0, -2)\)[/tex] and [tex]\(I (1, 2)\)[/tex]:
[tex]\[ K I = \sqrt{(1 - 0)^2 + (2 - (-2))^2} = \sqrt{1 + 16} = \sqrt{17} \][/tex]
Thus, [tex]\(K I = \sqrt{17}\)[/tex].
2. Calculate the distance [tex]\(K E\)[/tex]:
For vertices [tex]\(K (0, -2)\)[/tex] and [tex]\(E (4, -1)\)[/tex]:
[tex]\[ K E = \sqrt{(4 - 0)^2 + (-1 - (-2))^2} = \sqrt{16 + 1} = \sqrt{17} \][/tex]
Thus, [tex]\(K E = \sqrt{17}\)[/tex].
3. Calculate the distance [tex]\(I T\)[/tex]:
For vertices [tex]\(I (1, 2)\)[/tex] and [tex]\(T (7, 5)\)[/tex]:
[tex]\[ I T = \sqrt{(7 - 1)^2 + (5 - 2)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \][/tex]
Thus, [tex]\(I T = 3\sqrt{5}\)[/tex].
4. Calculate the distance [tex]\(T E\)[/tex]:
For vertices [tex]\(T (7, 5)\)[/tex] and [tex]\(E (4, -1)\)[/tex]:
[tex]\[ T E = \sqrt{(4 - 7)^2 + (-1 - 5)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \][/tex]
Thus, [tex]\(T E = 3\sqrt{5}\)[/tex].
5. Verification:
- We find that
[tex]\[ K I = K E = \sqrt{17} \][/tex]
and
[tex]\[ I T = T E = 3\sqrt{5} \][/tex]
- A kite is a quadrilateral with two pairs of adjacent sides that are equal. Here, [tex]\(K I = K E\)[/tex] and [tex]\(I T = T E\)[/tex].
Therefore, KITE is a kite because [tex]\(K I = K E\)[/tex] and [tex]\(I T = T E\)[/tex].
Step-by-Step Solution:
1. Calculate the distance [tex]\(K I\)[/tex]:
The distance formula is:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
For vertices [tex]\(K (0, -2)\)[/tex] and [tex]\(I (1, 2)\)[/tex]:
[tex]\[ K I = \sqrt{(1 - 0)^2 + (2 - (-2))^2} = \sqrt{1 + 16} = \sqrt{17} \][/tex]
Thus, [tex]\(K I = \sqrt{17}\)[/tex].
2. Calculate the distance [tex]\(K E\)[/tex]:
For vertices [tex]\(K (0, -2)\)[/tex] and [tex]\(E (4, -1)\)[/tex]:
[tex]\[ K E = \sqrt{(4 - 0)^2 + (-1 - (-2))^2} = \sqrt{16 + 1} = \sqrt{17} \][/tex]
Thus, [tex]\(K E = \sqrt{17}\)[/tex].
3. Calculate the distance [tex]\(I T\)[/tex]:
For vertices [tex]\(I (1, 2)\)[/tex] and [tex]\(T (7, 5)\)[/tex]:
[tex]\[ I T = \sqrt{(7 - 1)^2 + (5 - 2)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \][/tex]
Thus, [tex]\(I T = 3\sqrt{5}\)[/tex].
4. Calculate the distance [tex]\(T E\)[/tex]:
For vertices [tex]\(T (7, 5)\)[/tex] and [tex]\(E (4, -1)\)[/tex]:
[tex]\[ T E = \sqrt{(4 - 7)^2 + (-1 - 5)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \][/tex]
Thus, [tex]\(T E = 3\sqrt{5}\)[/tex].
5. Verification:
- We find that
[tex]\[ K I = K E = \sqrt{17} \][/tex]
and
[tex]\[ I T = T E = 3\sqrt{5} \][/tex]
- A kite is a quadrilateral with two pairs of adjacent sides that are equal. Here, [tex]\(K I = K E\)[/tex] and [tex]\(I T = T E\)[/tex].
Therefore, KITE is a kite because [tex]\(K I = K E\)[/tex] and [tex]\(I T = T E\)[/tex].
