/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1260 The circumcentre of the triangle... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 1260

The circumcentre of the triangle formed by the lines \(\mathrm{x}+\mathrm{y}=0, \mathrm{x}-\mathrm{y}=0\) and \(\mathrm{x}-7=0\) is \(\ldots \ldots\) (a) \((7,0)\) (b) \((3.5,0)\) (c) \((0,7)\) (d) \((3.5,3.5)\)

Short Answer

Expert verified
The circumcentre of the triangle formed by the lines $\mathrm{x}+\mathrm{y}=0, \mathrm{x}-\mathrm{y}=0$ and $\mathrm{x}-7=0$ is $(3.5, -3.5)$.

Step by step solution

01

Find the vertices of the triangle

To find the vertices, we need to solve the given linear equations in pairs: 1. x + y = 0 and x - y = 0 2. x + y = 0 and x - 7 = 0 3. x - y = 0 and x - 7 = 0 Solving the first pair of equations, we get: x = 0 and y = 0, which give us the first vertex A(0, 0). Solving the second pair of equations, we get: x = 7 and y = -7, which give us the second vertex B(7, -7). Solving the third pair of equations, we get: x = 7 and y = 7, which give us the third vertex C(7, 7).
02

Calculate the midpoints and slopes of two sides

Consider two sides AB and AC. Find the midpoints of these two sides: Midpoint of AB, M1 = \(\left(\frac{A_x + B_x}{2}, \frac{A_y + B_y}{2}\right)\) = \(\left(\frac{0+7}{2}, \frac{0-7}{2}\right)\) = \((3.5, -3.5)\) Midpoint of AC, M2 = \(\left(\frac{A_x + C_x}{2}, \frac{A_y + C_y}{2}\right)\) = \(\left(\frac{0+7}{2}, \frac{0+7}{2}\right)\) = \((3.5, 3.5)\) Find the slopes of the sides AB and AC: Slope of AB, m1 = \(\frac{B_y - A_y}{B_x - A_x}\) = \(\frac{-7-0}{7-0}\) = \(-1\) Slope of AC, m2 = \(\frac{C_y - A_y}{C_x - A_x}\) = \(\frac{7-0}{7-0}\) = \(1\)
03

Find the perpendicular bisectors of the two chosen sides

The slope of the line perpendicular to AB is the negative reciprocal of m1: m3 = \(\frac{1}{1}\) = \(1\). The slope of the line perpendicular to AC is the negative reciprocal of m2: m4 = \(\frac{1}{-1}\) = \(-1\). Now, we have the slopes and midpoints of the perpendicular bisectors M1R1 and M2R: Equation of M1R1: y - M1_y = m3 (x - M1_x) 鈬� y - (-3.5) = 1(x - 3.5) 鈬� y + 3.5 = x - 3.5 Equation of M2R: y - M2_y = m4 (x - M2_x) 鈬� y - 3.5 = -1(x - 3.5) 鈬� y - 3.5 = -x + 3.5
04

Calculate the intersection point of the two perpendicular bisectors

To find the circumcentre, we need to solve the two equations obtained in Step 3: 1. y + 3.5 = x - 3.5 2. y - 3.5 = -x + 3.5 Add both equations: 2y = -7 y = -3.5 Substitute the value of y in the first equation: -3.5 + 3.5 = x - 3.5 0 = x - 3.5 x = 3.5 Hence, the circumcentre is (3.5, -3.5), which is not among the given options.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Triangle Vertices
Understanding triangle vertices is crucial when identifying or analyzing the shape of a triangle. A vertex is simply a point where two or more lines meet, forming a corner. In the given exercise, the triangle is formed by three lines defined by the equations:
  • x + y = 0
  • x - y = 0
  • x - 7 = 0
By solving these equations in pairs, we can determine where each pair intersects, revealing the triangle's vertices:
  • The intersection of x + y = 0 and x - y = 0 yields the vertex A(0, 0).
  • The intersection of x + y = 0 and x - 7 = 0 gives the vertex B(7, -7).
  • The intersection of x - y = 0 and x - 7 = 0 results in the vertex C(7, 7).
Identifying these vertices is the first crucial step in understanding the geometry of the triangle.
Perpendicular Bisectors
Perpendicular bisectors are lines that perpendicularly cross another line segment at its midpoint. This concept is especially useful when finding a triangle's circumcenter, as the circumcenter is the point where the perpendicular bisectors of a triangle's sides intersect.
After identifying the triangle's vertices, we move to the calculation of its perpendicular bisectors.
For a line segment like AB or AC, the slope of the perpendicular bisector is the negative reciprocal of the original slope. For example:
  • The slope of AB is -1; thus, the perpendicular slope becomes 1.
  • The slope of AC is 1; thus, its perpendicular slope becomes -1.
With these perpendicular slopes in hand, we can construct the equations for these bisectors, using the midpoints we found earlier to find the triangle's circumcenter.
Midpoints of Sides
Midpoints play a vital role in understanding the properties of triangles, particularly when looking for the circumcenter. A midpoint is the point exactly halfway along a side of the triangle. To find it, you calculate the average of the coordinates of the endpoints.
In our example, we need these midpoints to determine the bisectors:
  • The midpoint of side AB is calculated as follows: \( \left( \frac{0+7}{2}, \frac{0-7}{2} \right) \), resulting in (3.5, -3.5).
  • Similarly, the midpoint of side AC is: \( \left( \frac{0+7}{2}, \frac{0+7}{2} \right) \), which is (3.5, 3.5).
These midpoints are critical as they help define the location for constructing the perpendicular bisectors, bringing us one step closer to locating the circumcenter.
Slopes of Lines
The slope of a line is a measure of its steepness and direction. In the context of triangles, understanding line slopes is essential for finding perpendicular bisectors. Slopes help determine how lines relate to one another:
  • Lines with slopes that are negative reciprocals are perpendicular to each other.
When dealing with a triangle, these relationships are essential for defining perpendicular bisectors which converge at the circumcenter.
  • For side AB, with coordinates (0,0) and (7,-7), the slope is calculated by \( \frac{-7}{7} = -1 \).
  • For side AC with coordinates (0,0) and (7,7), the slope is \( \frac{7}{7} = 1 \).
Using these slopes, we determine the perpendicular slopes required to find the bisectors necessary for locating the triangle's circumcenter. Understanding how to compute and apply these slopes is a key skill in geometry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equation of three sides of triangle are \(\mathrm{x}=2, \mathrm{y}+1=0\) and \(\mathrm{x}+2 \mathrm{y}=4\). The coordinates of the circumcentre of the triangle is (a) \((4,0)\) (b) \((2,-1)\) (c) \((0,4)\) (d) \((-1,2)\)

\(\mathrm{A}(1,0)\) and \(\mathrm{B}(-1,0)\), then the locus of points satisfying \(\mathrm{AQ}-\mathrm{BQ}=\pm 1\) is \(\ldots \ldots\) (a) \(12 \mathrm{x}^{2}+4 \mathrm{y}^{2}=3\) (b) \(12 \mathrm{x}^{2}-4 \mathrm{y}^{2}=3\) (c) \(12 \mathrm{x}^{2}-4 \mathrm{y}^{2}=-3\) (d) \(12 x^{2}+4 y^{2}=-3\)

A straight line passes through a point \(\mathrm{A}(1,2)\) and makes an angle \(60^{\circ}\) with the \(\mathrm{x}\) -axis. This line intersect the line \(\mathrm{x}+\mathrm{y}=6\) at \(\mathrm{P}\). Then AP will be (a) \(3(\sqrt{3}+1)\) (b) \(3(\sqrt{3}-1)\) (c) \((\sqrt{3}+1)\) (d) \(3 \sqrt{3}\)

The equation of line equidistant from the points \(\mathrm{A}(1,-2)\) and \(\mathrm{B}(3,4)\) and making congruent angles with the coordinate axes is \(\ldots \ldots \ldots\) (a) \(x+y=1\) (b) \(y-x+1=0\) (c) \(\mathrm{y}-\mathrm{x}-1=0\) (d) \(y-x=2\)

If the \(\mathrm{y}\) -intercept of the perpendicular bisector of the segment obtained by joining \(\mathrm{P}(1,4)\) and \(\mathrm{Q}(\mathrm{k}, 3)\) is \(-4\) then \(\mathrm{k}=\ldots \ldots\) (a) 1 (b) 2 (c) \(-2\) (d) \(-4\)
See all solutions

Recommended explanations on English Textbooks

Syntax

Read Explanation

Global English

Read Explanation

Free Response Essay

Read Explanation

Prosody

Read Explanation

Single Paragraph Essay

Read Explanation

Lexis and Semantics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.