Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 1347

Let PS be the median of the triangle with vertices \(\mathrm{P}(2,2)\), \(\mathrm{Q}(6,-1)\) and \(\mathrm{R}(7,3)\). The equation of the line passing through \((1,-1)\) and parallel to \(P S\) is (a) \(2 x-9 y-7=0\) (b) \(2 x-9 y-11=0\) (c) \(2 x+9 y-11=0\) (d) \(2 x+9 y+7=0\)

Short Answer

Expert verified
The short answer is: the equation of the line parallel to the median PS and passing through the point (1, -1) is (b) \(2x - 9y - 11 = 0\).

Step by step solution

01

Find the midpoint of QR.

The coordinates of points Q and R are given as (6,-1) and (7,3) respectively. We can find the midpoint S of QR using the midpoint formula: \[ S = \left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) = \left(\frac{6+7}{2}, \frac{-1+3}{2}\right) = (6.5, 1) \]
02

Calculate the slope of PS.

Now that we have the coordinates of the endpoints of the median PS, i.e., P(2,2) and S(6.5,1), we can find the slope of the line segment PS using the following formula: \[ m_{PS} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{1-2}{6.5-2} = \frac{-1}{4.5} = -\frac{2}{9} \]
03

Write the equation of a line parallel to PS and passing through (1,-1).

We have the slope of the line we need, \(-\frac{2}{9}\). Since this line passes through the point (1,-1), we can use the point-slope form for the equation of the line: \[ y - y_1 = m(x - x_1) \] Substituting the given point and the slope of the parallel line, we get: \[ y - (-1) = -\frac{2}{9}(x - 1) \] Simplifying the equation, we have: \[ 9(y + 1) = -2(x - 1) \] Expanding and rearranging this equation, we get the final equation as: \[ 2x - 9y - 11 = 0 \] So, the correct answer is (b) \(2x - 9y - 11 = 0\).

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

A square of side 'a' lies above the \(\mathrm{x}\) -axis and has one vertex at the origin. The side passing through the origin makes an angle o. \([0<\alpha<(\pi / 4)]\) with the positive direction of \(\mathrm{x}\) -axis. The equation of the diagonal not passing through the origin is (a) \(y(\cos \alpha-\sin \alpha)-x(\sin \alpha-\cos \alpha)=a\) (b) \(y(\cos \alpha+\sin \alpha)+x(\sin \alpha-\cos \alpha)=a\) (c) \(y(\cos \alpha-\sin \alpha)-x(\sin \alpha+\cos \alpha)=a\) (d) \(y(\cos \alpha+\sin \alpha)+x(\cos \alpha-\sin \alpha)=a\)

Let \(\mathrm{A}(2,-3)\) and \(\mathrm{B}(-2,1)\) be vertices of a triangle \(\mathrm{ABC}\). If the centroid of this triangle moves on the line \(2 x+3 y=1\), then locus of the vertex \(\mathrm{C}\) is the line (a) \(2 \mathrm{x}+3 \mathrm{y}=9\) (b) \(2 x-3 y=7\) (c) \(3 \mathrm{x}+2 \mathrm{y}=5\) (d) \(3 x-2 y=3\)

The in centre of a triangle whose vertices \(\mathrm{A}(2,4), \mathrm{B}(2,6)\) and \(\mathrm{C}(2+\sqrt{3}, 5)\) is.... (a) \([2+(1 / \sqrt{3}), 5]\) (b) \([1+\\{1 /(2 \sqrt{3})\\},(5 / 2)]\) (c) \((2,5)\) (d) None of these

If the \(\mathrm{x}\) -coordinate of the point of intersection of the lines \(3 \mathrm{x}+4 \mathrm{y}=9\) and \(\mathrm{y}=\mathrm{mx}+1\) is an integer, then the integer value of \(\mathrm{m}\) is \(\ldots \ldots\) (a) 2 (b) 0 (c) 4 (d) 1

A square of side a lies above the \(\mathrm{x}\) -axis and has one vertex at the origin. The side passing through the origin makes an angle a \(\alpha[0<\alpha<(\pi / 4)]\) with the positive direction of \(\mathrm{x}\) -axis. The eq. of its diagonal not passing through the origin is: (A) \(\mathrm{y}(\cos \alpha+\sin \alpha)+\mathrm{x}(\sin \alpha-\cos \alpha)=\mathrm{a}\) (B) \(\mathrm{y}(\cos \alpha+\sin \alpha)+\mathrm{x}(\sin \alpha+\cos \alpha)=\mathrm{a}\) (C) \(\mathrm{y}(\cos \alpha+\sin \alpha)+\mathrm{x}(\cos \alpha-\sin \alpha)=\mathrm{a}\) (D) \(y(\cos \alpha-\sin \alpha)-x(\sin \alpha-\cos \alpha)=a\)
See all solutions

Recommended explanations on English Textbooks

5 Paragraph Essay

Read Explanation

Prosody

Read Explanation

Cues and Conventions

Read Explanation

Sociolinguistics

Read Explanation

Email

Read Explanation

Listening and Speaking

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