Show that the points $A(-2, 1)$,$B(2, -2)$,and $C(5, 2)$ are the vertices of a right-angled triangle.
(N/A) Let the given points be $A(-2, 1)$,$B(2, -2)$,and $C(5, 2)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB^2 = (2 - (-2))^2 + (-2 - 1)^2 = (4)^2 + (-3)^2 = 16 + 9 = 25$.
$BC^2 = (5 - 2)^2 + (2 - (-2))^2 = (3)^2 + (4)^2 = 9 + 16 = 25$.
$AC^2 = (5 - (-2))^2 + (2 - 1)^2 = (7)^2 + (1)^2 = 49 + 1 = 50$.
We observe that $AB^2 + BC^2 = 25 + 25 = 50 = AC^2$.
Since the sum of the squares of two sides is equal to the square of the third side,by the converse of the Pythagoras theorem,$\triangle ABC$ is a right-angled triangle with the right angle at vertex $B$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB^2 = (2 - (-2))^2 + (-2 - 1)^2 = (4)^2 + (-3)^2 = 16 + 9 = 25$.
$BC^2 = (5 - 2)^2 + (2 - (-2))^2 = (3)^2 + (4)^2 = 9 + 16 = 25$.
$AC^2 = (5 - (-2))^2 + (2 - 1)^2 = (7)^2 + (1)^2 = 49 + 1 = 50$.
We observe that $AB^2 + BC^2 = 25 + 25 = 50 = AC^2$.
Since the sum of the squares of two sides is equal to the square of the third side,by the converse of the Pythagoras theorem,$\triangle ABC$ is a right-angled triangle with the right angle at vertex $B$.
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$1$. $\square ABCD$ is a rhombus. $a$. $\overline{AC}$ and $\overline{BD}$ bisect each other. $2$. $\square ABCD$ is a parallelogram. $b$. $\overline{AC}$ and $\overline{BD}$ bisect each other at right angles. $3$. $\square ABCD$ is a rectangle. $c$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other at right angles. $4$. $\square ABCD$ is a square. $d$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other.
| $1$. $\square ABCD$ is a rhombus. | $a$. $\overline{AC}$ and $\overline{BD}$ bisect each other. |
| $2$. $\square ABCD$ is a parallelogram. | $b$. $\overline{AC}$ and $\overline{BD}$ bisect each other at right angles. |
| $3$. $\square ABCD$ is a rectangle. | $c$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other at right angles. |
| $4$. $\square ABCD$ is a square. | $d$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other. |
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