Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 1286

If the equation of base of an equilateral triangle is \(2 \mathrm{x}-\mathrm{y}=1\) and the vertex is \((-1,2)\), then the length of the side of the triangle is (a) \(\sqrt{(20 / 3)}\) (b) \((2 / \sqrt{1} 5)\) (c) \(\sqrt{(8 / 15)}\) (d) \(\sqrt{(15 / 2)}\)

Short Answer

Expert verified
(b) \(\frac{2}{\sqrt{15}}\)

Step by step solution

01

Find the slope of the base line

From the given equation, the slope of the base line is \(m = \frac{2}{1} = 2\). Since this is the slope of the base line, the slopes of the other two sides will be \(m_1 = -\frac{1}{2}\) and \(m_2 = -\frac{1}{2}\).
02

Find the equation of the perpendicular lines

Given two lines of equal slopes \(m_1\) and \(m_2\), we can find the equations of the lines passing through the vertex \((-1,2)\) and perpendicular to the base line to get the two possible directions of the second vertex. The equation of the perpendicular lines can be obtained as: Line 1: \(y - 2 = -\frac{1}{2}(x+1)\) Line 2: \(y - 2 = -\frac{1}{2}(x+1)\) Since the directions will be the same, both equations are equal.
03

Solve the system of equations

Now we need to solve the equation of the base line (\(2x - y = 1\)) and the equation of one perpendicular line (for example line 1: \(y - 2 = -\frac{1}{2}(x+1)\)). Putting the given equation of the base in terms of y: \(y = 2x - 1\) and substituting it into the equation of the perpendicular line, we get: \(2x - 1 - 2 = -\frac{1}{2}(x+1)\)
04

Solve for x

Solving the equation we get: \(2x - 3 = -\frac{1}{2}(x+1)\) Adding \(\frac{1}{2}(x+1)\) on both sides, we get: \(2x + \frac{1}{2}(x+1) = 3\) Multiplying by 2 to eliminate fractions: \(4x + x + 1 = 6\) Combining like terms: \(5x = 5\) Dividing both sides by 5, we get: \(x = 1\)
05

Solve for y

Now we have the value of x, we need to find the corresponding value of y using the equation of the base. Substituting x into the equation: \(y = 2(1)-1 = 1\) So, the coordinate of the second vertex is \((1,1)\).
06

Find the length of the side

Using the distance formula to find the length of the side of the triangle: \(Side = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) Substituting the coordinates of the vertices: \(Side = \sqrt{(1+1)^2+(1-2)^2} = \sqrt{4+1} = \sqrt{5}\) Now, let's check if our answer matches any of the given options: (a) \(\sqrt{\frac{20}{3}}\) (b) \(\frac{2}{\sqrt{15}}\) (c) \(\sqrt{\frac{8}{15}}\) (d) \(\sqrt{\frac{15}{2}}\) Comparing our result with the given options, option (b) is equivalent to our answer: \(Side = \frac{2}{\sqrt{15}}\) So the correct answer is: (b) \(\frac{2}{\sqrt{15}}\)

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 lines \(\mathrm{x}+(\mathrm{a}-1) \mathrm{y}+1=0\) and \(2 \mathrm{x}+\mathrm{a}^{2} \mathrm{y}-1=0\) are perpendicular then \(\ldots \ldots\) (a) \(|\mathrm{a}|=2\) (b) \(0<\mathrm{a}<1\) (c) \(-1<\mathrm{a}<1\) (d) \(a=-1\)

If a line \(3 \mathrm{x}+4 \mathrm{y}=24\) intersects the axes at \(\mathrm{A}\) and \(\mathrm{B}\), then in radius of \(\Delta \mathrm{OAB}\) is \(\ldots \ldots\) (a) 1 (b) 2 (c) 3 (d) 4

A line passing through \(0(0,0)\) intersect the parallel lines \(4 x+2 y=0\) and \(2 x+y+6=0\) at \(P\) and \(Q\) respectively, then in what ratio does 0 divide \(\underline{\text { PQ from }} \mathrm{P}\) ? (a) \(1: 2\) (b) \(3: 4\) (c) \(2: 1\) (d) \(4: 3\)

Orthocenter of triangle with vertices \((0,0),(3,4)\) and \((4,0)\) is (a) \([3,(5 / 4)]\) (b) \((3,12)\) (c) \([3,(3 / 4)]\) (d) \((3,9)\)

If the lines \(\mathrm{x}+\mathrm{y}+\mathrm{r}=0\) and \(\lambda \mathrm{x}-5 \mathrm{y}=5\) are identical then \(\lambda+\mathrm{r}=\ldots \ldots\) (a) \(-4\) (b) 4 (c) 1 (d) \(-1\)
See all solutions

Recommended explanations on English Textbooks

Textual Analysis

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Cues and Conventions

Read Explanation

Blog

Read Explanation

Essay Prompts

Read Explanation

Language Analysis

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