Equation for a line through an origin

  • Thread starter Thread starter Poetria
  • Start date Start date
  • Tags Tags
    Line Origin
Click For Summary

Homework Help Overview

The discussion revolves around the equation of a line through the origin, specifically examining the expression \(-4y + x = 0\) and its interpretation in the context of an automatic grading system.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of the equation provided and the criteria used by the automatic grader. There is a discussion about the uniqueness of solutions to the equation form \(ay + bx = 0\) and the implications of normalization or positivity of coefficients.

Discussion Status

Some participants are exploring different forms of the equation and considering adjustments based on feedback from the grading system. There is an ongoing examination of the requirements for the expression and how they might affect the grading outcome.

Contextual Notes

Participants note that the grading system may have specific expectations regarding the form of the equation, such as requiring a positive coefficient for \(a\) or a normalized vector. This introduces uncertainty about the criteria for correctness.

Poetria
Messages
267
Reaction score
42
Homework Statement
Find the equation for the line through the origin that is perpendicular to the vector (-1,4). Enter the equation in the form: ##a*y+b*x=0##
Relevant Equations
Dot product of the vector (-1,4) and the vector parallel to the line has to be 0.
Why is this wrong?
$$-4*y+x=0$$

$$\vec (-1, 4)\cdot\vec (4,1)=0$$
 
Last edited:
Physics news on Phys.org
Why do you think it is wrong?
 
  • Like
Likes   Reactions: Poetria
It has been marked as wrong by an automatic grader. But I don't understand why. :(
 
Poetria said:
It has been marked as wrong by an automatic grader. But I don't understand why. :(
There is no unique solution to ##ay+bx=0## because all ##acy+acx=0 \;(c\neq 0)## are solutions, too. Could be that whatever checked it expected a normalized vector, or a positive ##a##.
 
  • Like
Likes   Reactions: Poetria
It is stated: expression in the form ##a*y+b*x##. I will try with the positive a then.

$$4*y-x=0$$

Many thanks. :)
 
  • Like
Likes   Reactions: berkeman

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
17
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
7
Views
2K
Replies
18
Views
2K