/*! 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 81 Consider the following pairs of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 81

Consider the following pairs of lines. Determine whether the lines are parallel or intersecting. If the lines intersect, then determine the point of intersection. a. \(x=1+s, y=2 s\) and \(x=1+2 t, y=3 t\) b. \(x=2+5 s, y=1+s\) and \(x=4+10 t, y=3+2 t\) c. \(x=1+3 s, y=4+2 s\) and \(x=4-3 t, y=6+4 t\)

Short Answer

Expert verified
Based on the given information, for each pair of lines: a. The lines are not parallel and intersect at point (5, 6). b. The lines are parallel and do not intersect. c. The lines are not parallel and intersect at point (4, 6).

Step by step solution

01

Check if the direction vectors are parallel

For lines \(x=1+s, y=2 s\) and \(x=1+2t, y=3t\), the direction vectors are \((1, 2)\) and \((2, 3)\). The lines are parallel if the direction vectors are proportional. Since \(\frac{2}{1} \neq \frac{3}{2}\), the direction vectors are not parallel, and thus, the lines are not parallel.
02

Solve for the intersection point

Since the lines are not parallel, they must intersect. We can find the intersection point by setting \(x\) and \(y\) equal in both equations: \((1+s) = (1+2t)\) and \((2s) = (3t)\) Solve the first equation for \(s\): \(s=2t-1\). Substitute this expression into the second equation: \(2(2t-1)=3t \implies t=2\). Now, substitute \(t=2\) back into either line's equation to find the intersection point: \(x=1+2(2)=5\), \(y=3(2)=6\). Thus, the point of intersection is \((5, 6)\). #b. Pair (b) #
03

Check if the direction vectors are parallel

For lines \(x=2+5s, y=1+s\) and \(x=4+10t, y=3+2t\), the direction vectors are \((5, 1)\) and \((10, 2)\). The lines are parallel if the direction vectors are proportional. Since \(\frac{10}{5} = \frac{2}{1}\), the direction vectors are parallel, and thus, the lines are parallel.
04

No intersection point

As the lines are parallel, there is no intersection point. #c. Pair (c) #
05

Check if the direction vectors are parallel

For lines \(x=1+3s, y=4+2s\) and \(x=4-3t, y=6+4t\), the direction vectors are \((3, 2)\) and \((-3, 4)\). The lines are parallel if the direction vectors are proportional. Since \(\frac{-3}{3} \neq \frac{4}{2}\), the direction vectors are not parallel, and thus, the lines are not parallel.
06

Solve for the intersection point

Since the lines are not parallel, they must intersect. We can find the intersection point by setting \(x\) and \(y\) equal in both equations: \((1+3s) = (4-3t)\) and \((4+2s) = (6+4t)\) Solve the first equation for \(s\): \(s = -t+1\). Substitute this expression into the second equation: \(4+2(-t+1)=6+4t \implies t=1\). Now, substitute \(t=1\) back into either line's equation to find the intersection point: \(x=1+3(1)=4\), \(y=4+2(1)=6\). Thus, the point of intersection is \((4, 6)\).

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!

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 butterfly curve of Example 8 is enhanced by adding a term: $$r=e^{\sin \theta}-2 \cos 4 \theta+\sin ^{5}(\theta / 12), \quad \text { for } 0 \leq \theta \leq 24 \pi$$ a. Graph the curve. b. Explain why the new term produces the observed effect. (Source: S. Wagon and E. Packel, Animating Calculus, Freeman, 1994)

A focal chord of a conic section is a line through a focus joining two points of the curve. The latus rectum is the focal chord perpendicular to the major axis of the conic. Prove the following properties. Let \(L\) be the latus rectum of the parabola \(y^{2}=4 p x,\) for \(p>0\) Let \(F\) be the focus of the parabola, \(P\) be any point on the parabola to the left of \(L,\) and \(D\) be the (shortest) distance between \(P\) and \(L\) Show that for all \(P, D+|F P|\) is a constant. Find the constant.

Show that the set of points equidistant from a circle and a line not passing through the circle is a parabola. Assume the circle, line, and parabola lie in the same plane.

Find parametric equations that describe the circular path of the following objects. Assume \((x, y)\) denotes the position of the object relative to the origin at the center of the circle. Use the units of time specified in the problem. There is more than one way to describe any circle. A Ferris wheel has a radius of \(20 \mathrm{m}\) and completes a revolution in the clockwise direction at constant speed in 3 min. Assume that \(x\) and \(y\) measure the horizontal and vertical positions of a seat on the Ferris wheel relative to a coordinate system whose origin is at the low point of the wheel. Assume the seat begins moving at the origin.

Find parametric equations for the following circles and give an interval for the parameter values. Graph the circle and find a description in terms of \(x\) and \(y .\) Answers are not unique. A circle centered at the origin with radius \(12,\) generated clockwise with initial point \((0,12)\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Decision Maths

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.