Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 1309

The equation of the lines with slope \(-2\) and intersecting \(\mathrm{x}\) -axis at points distance 3 unit from the origin is...... (a) \(2 \mathrm{x}+\mathrm{y} \pm 6=0\) (b) \(x+2 y \pm 6=0\) (c) \(2 \mathrm{x}+\mathrm{y} \pm 3=0\) (d) \(x+2 y \pm 3=0\)

Short Answer

Expert verified
The correct answer is (a) \(2x+y\pm6=0\).

Step by step solution

01

Find the points of intersection

The lines intersect the x-axis at points at a distance of ±3 units from the origin. The x-axis points are at: \((-3,0)\) and \((3,0)\).
02

Using the slope-point form

Since we know the slope, m, and a point on the line, we can use the slope-point form of the line equation: \[y-y_1=m(x-x_1)\] where \((x_1, y_1)\) is the point and \(m\) is the slope.
03

Write the equation for each point

For the point \((-3,0)\): \[y-0=(-2)(x-(-3))\] \[y=-2(x+3)\] For the point \((3,0)\): \[y-0=(-2)(x-3)\] \[y=-2(x-3)\]
04

Convert to standard form

Now, convert both equation to the standard form, i.e., Ax + By = C. For the line passing through \((-3,0)\): \[y=-2(x+3)\] \[y=-2x-6\] \[2x+y+6=0\] For the line passing through \((3,0)\): \[y=-2(x-3)\] \[y=-2x+6\] \[2x+y-6=0\]
05

Compare with options

Comparing these equations with the given options: \(2x+y+6=0\) matches option (a): \(2x+y\pm6=0\) and \(2x+y-6=0\) also matches option (a): \(2x+y\pm6=0\) So, the correct answer is: (a) \(2x+y\pm6=0\)

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

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

Line \(\mathrm{ax}+\mathrm{by}+\mathrm{p}=0\) makes angle \((\pi / 4)\) with \(x \cos \alpha+\) ysin \(\alpha=p, p \in R^{+} .\) If these lines and the line \(x \sin \alpha-y \cos \alpha=0\) are concurrent then (a) \(a^{2}+b^{2}=p^{2}\) (b) \(a^{2}+b^{2}=2 p^{2}\) (c) \(2\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)=\mathrm{p}\) (d) \(a^{2}-b^{2}=2 p^{2}\)

If two perpendicular lines passing through origin intersect the line \((\mathrm{x} / \mathrm{a})+(\mathrm{y} / \mathrm{b})=1, \mathrm{a} \neq 0, \mathrm{~b} \neq 0\) at \(\mathrm{A}\) and \(\mathrm{B}\), then \(\left\\{1 /(\mathrm{OA})^{2}\right\\}+\left\\{1 /(\mathrm{OB})^{2}\right\\}=\ldots \ldots\) (a) \(\left(1 / \mathrm{a}^{2}\right)-\left(1 / \mathrm{b}^{2}\right)\) (b) \(\left[(\mathrm{ab}) /\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)\right]\) (c) \(\left[\left(a^{2}+b^{2}\right) /\left(a^{2} b^{2}\right)\right]\) (d) None of these

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) \(y(\cos \alpha+\sin \alpha)+x(\cos \alpha-\sin \alpha)=a\)

A variable straight line passes through a fixed point \((a, b)\) intersecting the coordinate axes at \(\mathrm{A}\) and \(\mathrm{B}\). If ' \(\mathrm{O}^{\prime}\) is the origin, then the locus of the centroid of the triangle \(\mathrm{OAB}\) is (a) \(b x+a y=3 x y\) (b) \(\mathrm{bx}+\mathrm{ay}=2 \mathrm{xy}\) (c) \(a x+b y=3 x y\) (d) \(a x+b y=2 x y\)
See all solutions

Recommended explanations on English Textbooks

Textual Analysis

Read Explanation

Free Response Essay

Read Explanation

Essay Plan

Read Explanation

Graphology

Read Explanation

Syntax

Read Explanation

Phonetics

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