Question

which term describes the figure that is formed by the following coordinates? a(-4,3), b(2,3), c(3,-1), and d(-5,-1)

Explanation:

Step1: Calculate side lengths

Use distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
\(AB\): \(x_1=-4,y_1=3;x_2=2,y_2=3\)
\(AB=\sqrt{(2 - (-4))^2 + (3 - 3)^2}=\sqrt{6^2}=6\)
\(BC\): \(x_1=2,y_1=3;x_2=3,y_2=-1\)
\(BC=\sqrt{(3 - 2)^2 + (-1 - 3)^2}=\sqrt{1 + 16}=\sqrt{17}\)
\(CD\): \(x_1=3,y_1=-1;x_2=-5,y_2=-1\)
\(CD=\sqrt{(-5 - 3)^2 + (-1 - (-1))^2}=\sqrt{(-8)^2}=8\)
\(DA\): \(x_1=-5,y_1=-1;x_2=-4,y_2=3\)
\(DA=\sqrt{(-4 - (-5))^2 + (3 - (-1))^2}=\sqrt{1 + 16}=\sqrt{17}\)

Step2: Check slopes for parallelism

Slope formula: \(m=\frac{y_2 - y_1}{x_2 - x_1}\)
Slope of \(AB\): \(\frac{3 - 3}{2 - (-4)} = 0\) (horizontal line)
Slope of \(CD\): \(\frac{-1 - (-1)}{-5 - 3} = 0\) (horizontal line, so \(AB \parallel CD\))
Slope of \(BC\): \(\frac{-1 - 3}{3 - 2} = -4\)
Slope of \(DA\): \(\frac{3 - (-1)}{-4 - (-5)} = 4\)? Wait, no: \(\frac{3 - (-1)}{-4 - (-5)}=\frac{4}{1}=4\)? Wait, earlier \(DA\) length was \(\sqrt{17}\), \(BC\) too. Wait, no, recalculate \(DA\) slope: \(x_1=-5,y_1=-1;x_2=-4,y_2=3\): \(\frac{3 - (-1)}{-4 - (-5)}=\frac{4}{1}=4\). \(BC\) slope: \(\frac{-1 - 3}{3 - 2}=\frac{-4}{1}=-4\). Wait, maybe I mixed up points. Wait, the points are \(A(-4,3)\), \(B(2,3)\), \(C(3,-1)\), \(D(-5,-1)\). So \(DA\) is from \(D(-5,-1)\) to \(A(-4,3)\): slope is \(\frac{3 - (-1)}{-4 - (-5)} = 4\). \(BC\) is from \(B(2,3)\) to \(C(3,-1)\): slope is \(\frac{-1 - 3}{3 - 2} = -4\). Wait, but \(AB\) and \(CD\) are horizontal (slope 0), so they are parallel. \(BC\) and \(DA\): wait, no, maybe I made a mistake in point order. Let's check \(AD\) and \(BC\). Wait, maybe the figure is a trapezoid? Wait, no, trapezoid has one pair of parallel sides. But \(AB\) and \(CD\) are parallel (both horizontal, same y - coordinate for \(AB\) (y = 3) and \(CD\) (y = -1)). Now check if it's a trapezoid (one pair) or parallelogram (two pairs). Wait, slope of \(AB\) is 0, slope of \(CD\) is 0 (parallel). Slope of \(BC\): from \(B(2,3)\) to \(C(3,-1)\): \(\frac{-1 - 3}{3 - 2}=-4\). Slope of \(AD\): from \(A(-4,3)\) to \(D(-5,-1)\): \(\frac{-1 - 3}{-5 - (-4)}=\frac{-4}{-1}=4\). Wait, that's not parallel. Wait, maybe the points are \(A(-4,3)\), \(B(2,3)\), \(C(3,-1)\), \(D(-5,-1)\). Let's check \(AB\) length: 6, \(CD\) length: 8 (wait, earlier I thought \(CD\) was 8? Wait, \(x_1=3,y_1=-1;x_2=-5,y_2=-1\): distance is \(|-5 - 3| = 8\) (since y - coordinates same). \(AB\) is \(|2 - (-4)| = 6\) (y same). So \(AB\) and \(CD\) are parallel (horizontal) but different lengths. \(BC\) and \(AD\): length of \(BC\): \(\sqrt{(3 - 2)^2 + (-1 - 3)^2}=\sqrt{1 + 16}=\sqrt{17}\). Length of \(AD\): \(\sqrt{(-4 - (-5))^2 + (3 - (-1))^2}=\sqrt{1 + 16}=\sqrt{17}\). So \(BC = AD\), and \(AB \parallel CD\) (one pair of parallel sides, non - parallel sides equal in length: isosceles trapezoid? Wait, but the options: trapezoid, parallelogram, rectangle, rhombus. Parallelogram needs two pairs of parallel sides. Rectangle needs right angles (slopes product - 1 for adjacent sides). Let's check angles. Slope of \(AB\) is 0, slope of \(BC\) is -4. Product is 0*(-4)=0≠ - 1, so not right angle. So it's a trapezoid (one pair of parallel sides). Wait, but maybe I messed up the points. Wait, the problem says "the figure that is represented by the following coordinates: A(-4,3), B(2,3), C(3,-1), D(-5,-1)". Let's plot mentally: A(-4,3), B(2,3) (horizontal line, length 6). C(3,-1), D(-5,-1) (horizontal line, length 8). Then B to C: (2,3) to (3,-1): right 1, down 4. D to A: (-5,-1) to (-4,3): right 1, up 4. So \(BC\) and \(AD\) are congruent and parallel? Wait, slope of \(BC\): \(\frac{-1 - 3}{3 - 2}=-4\), slope of \(AD\): \(\frac{3 - (-1)}{-4 - (-5)} = 4\). Wait, no, that's not parallel. Wait, maybe the correct pair is \(AB \parallel CD\) (both horizontal), so it's a trapezoid. So the answer should be trapezoid.

Answer:

a) trapezoid