Math Problem Statement

The points Nleft parenthesis, 5, comma, 5, right parenthesis(5,5), Oleft parenthesis, 3, comma, minus, 4, right parenthesis(3,−4), and Pleft parenthesis, minus, 2, comma, minus, 1, right parenthesis(−2,−1) form a triangle. Plot the points then click the "Graph Triangle" button., On this question page, you will first graph points on a coordinate grid. Following the graph are two buttons, "Graph Triangle" will drawn lines between your points, and "Reset" will remove all of your points. Once you press Graph Triangle, if your points are correct, the answer area will appear. Click on the graph to plot a point. Click a point to delete it. x y N O P Find the desired slopes and lengths, then fill in the words that characterize the triangle. Answer Attempt 1 out of 2

slope of NO = slope of OP = slope of PN = length of NO = length of OP = length of PN = Triangle NOP is    Submit Answer  

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Solution

To solve this problem and characterize the triangle formed by points N(5,5)N(5, 5), O(3,4)O(3, -4), and P(2,1)P(-2, -1), we need to find the slopes of the sides NONO, OPOP, and PNPN, as well as their lengths. Here is the solution step by step:


Step 1: Find the slopes

The slope of a line between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by: m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}

Slope of NONO:

mNO=4535=92=4.5m_{NO} = \frac{-4 - 5}{3 - 5} = \frac{-9}{-2} = 4.5

Slope of OPOP:

mOP=1(4)23=35=0.6m_{OP} = \frac{-1 - (-4)}{-2 - 3} = \frac{3}{-5} = -0.6

Slope of PNPN:

mPN=5(1)5(2)=670.857m_{PN} = \frac{5 - (-1)}{5 - (-2)} = \frac{6}{7} \approx 0.857


Step 2: Find the lengths

The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Length of NONO:

dNO=(35)2+(45)2=(2)2+(9)2=4+81=859.22d_{NO} = \sqrt{(3 - 5)^2 + (-4 - 5)^2} = \sqrt{(-2)^2 + (-9)^2} = \sqrt{4 + 81} = \sqrt{85} \approx 9.22

Length of OPOP:

dOP=(23)2+(1(4))2=(5)2+(3)2=25+9=345.83d_{OP} = \sqrt{(-2 - 3)^2 + (-1 - (-4))^2} = \sqrt{(-5)^2 + (3)^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83

Length of PNPN:

dPN=(5(2))2+(5(1))2=(7)2+(6)2=49+36=859.22d_{PN} = \sqrt{(5 - (-2))^2 + (5 - (-1))^2} = \sqrt{(7)^2 + (6)^2} = \sqrt{49 + 36} = \sqrt{85} \approx 9.22


Step 3: Characterize the triangle

To classify the triangle, compare slopes and lengths:

  1. Slopes of NONO, OPOP, and PNPN are not equal, so the triangle is not isosceles.
  2. Lengths NO=PN9.22NO = PN \approx 9.22, but OPNOOP \neq NO, so the triangle is isosceles.

Final Answer:

  1. Slope of NONO: 4.54.5
  2. Slope of OPOP: 0.6-0.6
  3. Slope of PNPN: 0.8570.857
  4. Length of NONO: 9.229.22
  5. Length of OPOP: 5.835.83
  6. Length of PNPN: 9.229.22
  7. Triangle NOPNOP is isosceles.

Would you like a plot of this triangle or further clarification?


Five Related Questions:

  1. How do you determine if a triangle is right, acute, or obtuse using slopes and lengths?
  2. How can the midpoint formula be applied to triangle geometry problems?
  3. What is the significance of perpendicular slopes in characterizing triangles?
  4. How do you prove that two sides of a triangle are equal using algebra?
  5. How can you find the area of a triangle using the vertices' coordinates?

Tip: When comparing slopes, remember perpendicular lines have slopes that are negative reciprocals.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Slopes of Lines
Distance Between Two Points
Triangle Classification

Formulas

Slope formula: m = (y2 - y1) / (x2 - x1)
Distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Classification of Triangles by Side Lengths
Slope and Parallel/Perpendicular Line Properties

Suitable Grade Level

Grades 9-11