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Chapter 1: Problem 87

By calculating the lengths of its sides, show that the triangle with vertices at the points \(A(1,2), B(5,5),\) and \(C(4,-2)\) is isosceles but not equilateral.

Short Answer

Expert verified
The triangle is isosceles with sides AB = 5, BC = 5鈭�2, CA = 5.

Step by step solution

01

Determine Length of Side AB

Using the distance formula, calculate the length of side AB. The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \[\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]For points A(1,2) and B(5,5), the calculation is:\[\sqrt{(5-1)^2 + (5-2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\]Thus, the length of AB is 5 units.
02

Determine Length of Side BC

We apply the distance formula to find the length of side BC. For points B(5,5) and C(4,-2), calculate as follows:\[\sqrt{(4-5)^2 + (-2-5)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}\]The length of the side BC is therefore \(5\sqrt{2}\) units.
03

Determine Length of Side CA

Calculate the distance for side CA with the distance formula. For points C(4,-2) and A(1,2):\[\sqrt{(1-4)^2 + (2+2)^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]Thus, the length of side CA is 5 units.
04

Identify Type of Triangle

Compare the calculated side lengths to determine the triangle type:- AB = 5- BC = \(5\sqrt{2}\)- CA = 5Since two sides (AB and CA) are equal, and one side (BC) is unequal, the triangle is isosceles. As none of the sides are all equal, it is not equilateral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance Formula
The distance formula is a fundamental tool in geometry. It allows us to calculate the precise distance between two points in a Cartesian plane. This formula derives from the Pythagorean theorem and is expressed as: \[\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\] where \(x_1, y_1\) and \(x_2, y_2\) represent the coordinates of the two points.

This formula is particularly useful because:
  • It provides a method to compute the straight-line distance between any two points, which is necessary for further geometric calculations.
  • It is used extensively in problems involving triangles, polygons, and circles.
To calculate the distances for the given problem, we applied this formula to different pairs of points: \(A(1,2), B(5,5),\text{and } C(4,-2)\). This enabled us to find the exact lengths of each side of the triangle, which is crucial for its classification.
Triangle Classification
Triangles can be classified based on their side lengths or angles. In this exercise, the classification is based on side lengths. Here are the key types:
  • Equilateral Triangle: All sides are equal.
  • Isosceles Triangle: Two sides are equal in length.
  • Scalene Triangle: All sides are different lengths.

In the provided problem, after calculating the side lengths using the distance formula, we identified that two sides were equal (AB = 5 units and CA = 5 units), and one side was different (BC = \(5\sqrt{2}\) units). Therefore, the triangle is isosceles.

It's important to recognize that an isosceles triangle always has two equal sides and can help simplify many geometric problems by offering symmetry, which makes it easier to apply other geometric rules and theorems.
Geometry Problem-Solving
In geometry problem-solving, understanding the properties and relationships within shapes is critical. Here鈥檚 how we applied this in classifying the triangle in this exercise:
  • Use of the Distance Formula: Calculating sides accurately provides the foundation for further analysis, like classification.
  • Comparison of Side Lengths: Critical for determining the type of triangle, especially in distinguishing between isosceles, equilateral, or scalene.
  • Logical Deduction: By comparing the calculated lengths, we deduced the triangle's type methodically and confirmed it complies with the characteristics of an isosceles triangle.


Effective problem-solving also involves checking the solution at each step to ensure accuracy. Applying systematic approaches like these increases accuracy and deepens comprehension of geometric principles. Encouraging consistency in using these processes will not only bolster understanding but also enhance overall mathematical ability.

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Most popular questions from this chapter

For $$f(x)=A \sin \left(\frac{2 \pi}{B}(x-C)\right)+D$$ identify \(A, B, C,\) and \(D\) for the sine functions in Exercises \(61-64\) and sketch their graphs (see Figure \(1.76 ) .\) $$ y=2 \sin (x+\pi)-1 $$

Industrial costs Dayton Power and Light, Inc., has a power plant on the Miami River where the river is 800 ft wide. To lay a new cable from the plant to a location in the city 2 mi downstream on the opposite side costs \(\$ 180\) per foot across the river and \(\$ 100\) per foot along the land. Graph cannot copy a. Suppose that the cable goes from the plant to a point \(Q\) on the opposite side that is \(x\) ft from the point \(P\) directly opposite the plant. Write a function \(C(x)\) that gives the cost of laying the cable in terms of the distance \(x\) . b. Generate a table of values to determine if the least expensive location for point \(Q\) is less than 2000 \(\mathrm{ft}\) or greater than 2000 \(\mathrm{ft}\) from point \(P .\)

The approximation sin \(x \approx x\) It is often useful to know that, when \(x\) is measured in radians, \(\sin x \approx x\) for numerically small values of \(x\) . In Section \(3.8,\) we will see why the approximation holds. The approximation error is less than 1 in 5000 if \(|x|<0.1\) a. With your grapher in radian mode, graph \(y=\sin x\) and \(y=x\) together in a viewing window about the origin. What do you see happening as \(x\) nears the origin? b. With your grapher in degree mode, graph \(y=\sin x\) and \(y=x\) together about the origin again. How is the picture different from the one obtained with radian mode? c. A quick radian mode check Is your calculator in radian mode? Evaluate \(\sin x\) at a value of \(x\) near the origin, say \(x=0.1\) . If \(\sin x \approx x\) , the calculator is in radian mode; if not, it isn't. Try it.

In Exercises \(1-6,\) find the domain and range of each function. $$ f(x)=1-\sqrt{x} $$

Graph the functions in Exercises \(23-26\) $$ f(x)=\left\\{\begin{array}{ll}{x,} & {0 \leq x \leq 1} \\ {2-x,} & {1
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