Question

2
select the correct answer from each drop-down menu.
the coordinates of a quadrilateral are (2,1), (-1,3), (-5,-3), and (-2,-5).
the quadrilateral is a because are parallel and since the product of the slopes of both pairs of segments is -1, .

Explanation:

Step1: Calculate slopes of sides

Let the points be \( A(2,1) \), \( B(-1,3) \), \( C(-5,-3) \), \( D(-2,-5) \).
Slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
Slope of \( AB \): \( \frac{3 - 1}{-1 - 2} = \frac{2}{-3} = -\frac{2}{3} \).
Slope of \( BC \): \( \frac{-3 - 3}{-5 - (-1)} = \frac{-6}{-4} = \frac{3}{2} \).
Slope of \( CD \): \( \frac{-5 - (-3)}{-2 - (-5)} = \frac{-2}{3} = -\frac{2}{3} \).
Slope of \( DA \): \( \frac{1 - (-5)}{2 - (-2)} = \frac{6}{4} = \frac{3}{2} \).

Step2: Identify parallel sides

\( AB \) (slope \( -\frac{2}{3} \)) and \( CD \) (slope \( -\frac{2}{3} \)) are parallel.
\( BC \) (slope \( \frac{3}{2} \)) and \( DA \) (slope \( \frac{3}{2} \)) are parallel.

Step3: Check perpendicularity (product of slopes)

Product of \( AB \) and \( BC \) slopes: \( -\frac{2}{3} \times \frac{3}{2} = -1 \).
Product of \( CD \) and \( DA \) slopes: \( -\frac{2}{3} \times \frac{3}{2} = -1 \).
So adjacent sides are perpendicular. A quadrilateral with two pairs of parallel sides (parallelogram) and perpendicular adjacent sides is a rectangle? Wait, no—wait, slope product -1 means perpendicular. Wait, actually, a parallelogram with perpendicular sides is a rectangle? Wait, no, wait: if both pairs of opposite sides are parallel (so parallelogram) and adjacent sides are perpendicular (slope product -1), then it's a rectangle? Wait, no, wait—wait, the slopes of \( AB \) and \( BC \) are \( -\frac{2}{3} \) and \( \frac{3}{2} \), which are negative reciprocals (product -1), so they are perpendicular. Similarly, \( BC \) and \( CD \): \( \frac{3}{2} \) and \( -\frac{2}{3} \), product -1. Wait, but actually, in a parallelogram, if adjacent sides are perpendicular, it's a rectangle. But wait, let's recheck:

Wait, the quadrilateral has vertices in order? Let's assume the order is \( A(2,1) \), \( B(-1,3) \), \( C(-5,-3) \), \( D(-2,-5) \), back to \( A \). So sides \( AB \), \( BC \), \( CD \), \( DA \).

So \( AB \parallel CD \), \( BC \parallel DA \) (opposite sides parallel: parallelogram). Then, check angles: product of slopes of \( AB \) and \( BC \) is -1, so they are perpendicular. So all angles are right angles. So it's a rectangle? Wait, but wait, maybe a square? No, lengths: let's check length of \( AB \): \( \sqrt{(-1-2)^2 + (3-1)^2} = \sqrt{9 + 4} = \sqrt{13} \). Length of \( BC \): \( \sqrt{(-5+1)^2 + (-3-3)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \). So not square, but rectangle (since opposite sides equal and parallel, and angles 90°). Wait, but the problem says "since the product of the slopes of both pairs of segments is -1"—wait, maybe the pairs are adjacent? Wait, the problem's drop-down: "because [first pair] are parallel and since the product of the slopes of both pairs of segments is -1, [conclusion]".

Wait, let's re-express:

First, find which sides are parallel: \( AB \) and \( CD \) (slope \( -\frac{2}{3} \)), \( BC \) and \( DA \) (slope \( \frac{3}{2} \)). So two pairs of opposite sides are parallel (so parallelogram). Then, the product of slopes of each pair of adjacent sides (e.g., \( AB \) and \( BC \), \( BC \) and \( CD \), etc.) is -1, meaning adjacent sides are perpendicular. So the quadrilateral is a **rectangle**? Wait, no, wait—actually, a parallelogram with perpendicular adjacent sides is a rectangle. But wait, let's confirm:

Wait, the first drop-down: options might be "rectangle", "rhombus", "square", "parallelogram", etc. Let's think again.

Wait, the slopes: \( AB \) and \( CD \) are parallel (same slope), \( BC \) and \( DA \) are parallel (same slope). So it's a parallelogram. Then, the product of the slopes of \( AB \) and \( BC \) is -1 (perpendicular), and product of \( BC \) and \( CD \) is -1 (perpendicular), etc. So all angles are right angles. So the parallelogram is a rectangle.

So the first blank: "rectangle" (or maybe "square"? But sides: \( AB \) length \( \sqrt{(-3)^2 + 2^2} = \sqrt{13} \), \( BC \) length \( \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \), so not square, so rectangle).

Second blank: "both pairs of opposite sides" (since \( AB \parallel CD \) and \( BC \parallel DA \)).

Third blank: "the angles are right angles" (since perpendicular sides form right angles).

Answer:

First drop - down: rectangle
Second drop - down: both pairs of opposite sides
Third drop - down: the angles are right angles