Log out Go to App
Go to App

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) ⇒ y - (-3.5) = 1(x - 3.5) ⇒ y + 3.5 = x - 3.5 Equation of M2R: y - M2_y = m4 (x - M2_x) ⇒ y - 3.5 = -1(x - 3.5) ⇒ 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

If the lengths of perpendicular drawn from the origin to the lines \(x \cos \alpha-y \sin \alpha=\sin 2 a \alpha\) and \(x \sin \alpha+y \cos \alpha=\cos 2 \alpha\) are \(p\) and \(q\) respectively, then \(p^{2}+q^{2}=\ldots \ldots\) (a) 4 (b) 3 (c) 2 (d) 1

If the point \([1+(t / \sqrt{2}), 2+(t / \sqrt{2})]\) lies between the two parallel lines \(x+2 y=1\) and \(2 x+4 y=15\), then the range of \(t\) is \(\ldots \ldots\) (a) \(0

If \(5 \mathrm{x}+12 \mathrm{y}+13=0\) is transformed into \(x \cos \alpha+\mathrm{ysin} \alpha=\mathrm{p}\) then \(\alpha=? \alpha \in[-\pi, \pi]\) (a) \(\cos ^{-1}[-(5 / 13)]\) (b) \(\sin ^{-1}[-(12 / 13)]\) (c) \(\tan ^{-1}(12 / 5)-\pi\) (d) \(\tan ^{-1}(12 / 5)\)

The equation of line passing through the point \((-5,4)\) and making the intercept of length \((2 / \sqrt{5})\) between the lines \(x+2 y-1=0\) and \(x+2 y+1=0\) is \(\ldots \ldots\) (a) \(2 \mathrm{x}-\mathrm{y}+4=0\) (b) \(2 x-y-14=0\) (c) \(2 \mathrm{x}-\mathrm{y}+14=0\) (d) None of these

The straight line \(7 \mathrm{x}-2 \mathrm{y}+10=0\) and \(7 \mathrm{x}+2 \mathrm{y}-10=0\) forms an isosceles triangle with the line \(\mathrm{y}=2\). Area of the triangle is equal to : (a) \((15 / 7)\) sq unit (b) \((10 / 7)\) sq unit (c) \((18 / 7)\) sq unit (d) \((10 / 13)\) sq unit
See all solutions

Recommended explanations on English Textbooks

Phonology

Read Explanation

English Grammar

Read Explanation

Phonetics

Read Explanation

Listening and Speaking

Read Explanation

Lexis and Semantics Summary

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.

Sign-up for free