Respuesta :
Answer:
#1) d. ΔJKL is not a right triangle because no two of its sides are perpendicular; #2) -1/3, 3, -7, is, two of these slopes have a product of -1; #3) a. Quadrilateral DEFG is a rhombus because opposite sides are parallel and all four sides have the same length; #4) 1, -1/6, 1, -2/5, is not, only one pair of opposite sides is parallel; #5) c. Quadrilateral PQRS is not a rectangle because it has only one right angle.
Step-by-step explanation:
#1) The slope of any line segment is found using the formula
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
For JK, this gives us (1-1)/(-5-0) = 0/-5 = 0. For KL this gives us (1--5)/(0-2) = 6/-2 = -3. For LJ this gives us (-5-1)/(2--5) = -6/7. None of these slopes are negative reciprocals, so none of the angles are right angles and this is not a right triangle.
#2) The slope of JK is (2-1)/(0-3) = 1/-3 = -1/3. The slope of KL is (1--5)/(3-1) = 6/2 = 3. The slope of LJ is (2--5)/(0-1) = 7/-1 = -7. Two of these slopes have a product of -1, 3 and -1/3. This means they are negative reciprocals so this has a right angle; this means JKL is a right triangle.
#3) The slope of DE is (5-4)/(-2-2) = 1/-4 = -1/4. The slope of EF is (4-0)/(2-0) = 4/2 = 2. The slope of FG is (0-1)/(0--4) = -1/4. The slope of GD is (1-5)/(-4--2) = -4/-2 = 2. Opposite sides have the same slope so they are parallel.
Next we use the distance formula to find the length of each side:
[tex]d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]
Using our points, the length of DE is
[tex]\sqrt{(5-4)^2+(-2-2)^2}=\sqrt{1^2+(-4)^2}=\sqrt{1+16}=\sqrt{17}[/tex]
The length of EF is
[tex]d=\sqrt{(4-0)^2+(2-0)^2}=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}[/tex]
The length of FG is
[tex]d=\sqrt{(0-1)^2+(0--4)^2}=\sqrt{(-1)^2+(4)^2}=\sqrt{1+16}=\sqrt{17}[/tex]
The length of GD is
[tex]d=\sqrt{(1-5)^2+(-4--2)^2}=\sqrt{(-4)^2+(-2)^2}=\sqrt{16+4}=\sqrt{20}[/tex]
Opposite sides have the same length and are parallel, so this is a parallelogram.
#4) The slope of AB is (-1-2)/(-4--1) = -3/-3 = 1. The slope of BC is (2-1)/(-1-5) = 1/-6 = -1/6. The slope of CD is (1--3)/(5-1) = 4/4 = 1. The slope of DA is (-3--1)/(1--4) = -2/5. Only one pair of opposite sides is parallel, so this is not a parallelogram.
#5) The slope of PQ is (2-4)/(-4-3) = -2/-7 = 2/7. The slope of QR is (4-0)/(3-5) = 4/-2 = -2. The slope of RS is (0--2)/(5--3) = 2/8 = 1/4. The slope of SP is (-2-2)/(-3--4) = -4/1 = -4. Only one pair of sides has slopes that are negative reciprocals; this means this figure only has 1 right angle, so it is not a rectangle.