Finding the Equation of a Line

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Discussion Overview

The discussion revolves around finding the equation of a line, specifically the perpendicular bisector of a line segment defined by two points on a circle. Participants explore the calculations of gradients, midpoints, and the relationship between different forms of the line equation, including slope-intercept and parametric forms.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant, Casio, calculates the gradient of line AB as -2/3 and finds the midpoint as [4/2, -2/2].
  • Casio proposes the equation of the perpendicular bisector as y = 2/3x - 6/2, based on the calculated gradient of the perpendicular line.
  • Casio expresses confusion regarding the equivalence of this equation to a parametric equation y = (3x - 8) / (2), noting discrepancies in the resulting y-values.
  • Another participant requests clarification on the original problem, indicating that the provided equation is in slope-intercept form rather than parametric form.
  • There is a correction regarding the gradient of the perpendicular line, with some participants stating it should be 3/2 instead of 2/3.
  • Casio acknowledges a typo regarding the gradient and reiterates the calculations for the midpoint and parametric equation.

Areas of Agreement / Disagreement

Participants do not reach consensus on the correct form of the line equation or the interpretation of the parametric equation. There are conflicting views on the gradient of the perpendicular line, and the discussion remains unresolved regarding the equivalence of the different forms of the line equation.

Contextual Notes

There are limitations in the clarity of the original problem statement, and some assumptions about the definitions of the equations are not explicitly stated. The discussion also reflects uncertainty in the calculations and interpretations of the equations involved.

Casio1
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OK I feel that I am stumped with this now. I have a circle problem. Given some coordinates I have calculated the gradient at -2/3. Then I am looking to find the midpoints, of the line, which are [4/2, -2/2]. I have then calculated the perpendicular of the line AB, which had a gradient - 2/3. The perdendicular gradient is 2/3.

I am then trying to find the equation of the line. This is what I have completed.

Bisector of line AB

bAB : y - [-2/2] = 2/3(x - 4/2) Implies y = 2/3x - 4/2 - 2/2 = y = 2/3x - 6/2

If this is the correct equation then I am at a loss because I am lead to believe that the parametric equation;

y = (3x - 8) / (2)

should give the same answer?

The parametric equation gives; y = (3(4/2) - 8) / (2) = - 2/2

I am quite OK with this because the midpoint is the same y value, so I think I got that right, but;

y = 2/3x - 6/2

Should give the same answer but to me does not?

y = 2/3(4/2) - 6/2 = - 1 2/3 which is - 1.67 (2dp)

The solutions are close but not exact, so I must be making a mistake somewhere I think?

Kind regards

Casio:confused:
 
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Could you please state the original problem verbatim?
 
OK, its going to be a long day.

I have sone points which represent a circle. A(5, -3), B(-1,1) and C(0,2)

I am asked to find the slope AB, this is - 2/3

I am asked to find the coordinates of the midpoint of the line segment AB, these are [4/2, - 2/2]

using my answers above I am asked to find the equation of the perpendicular bisector AB.I got; y = 2/3(x - 6/2)

Then I was asked to eliminate t from a parametric equation to show the same line found in above from finding the equation.

parametric equation; y = (3x - 8) / (2)

I am not sure which values of x I am suppose to enter to find the value of y in each case, but the closest I can get is using the parametric equation is -2/2 and using the other equation the closest I get is - 1.67.

They are not the same and at this point am a bit lost with it?

Kind regards

Casio
 
Casio said:
Then I was asked to eliminate t from a parametric equation to show the same line found in above from finding the equation.

parametric equation; y = (3x - 8) / (2)

Hi Casio, :)

I don't understand the above quoted statement. What you have given is the slope intercept form of a line not the parametric form.

Maybe you would find the following video at Khan Academy useful in understanding parametric equations.

Linear Algebra: Parametric Representations of Lines | Linear Algebra | Khan Academy

Kind Regards,
Sudharaka.
 
OK I feel that I am stumped with this now. I have a circle problem. Given some coordinates I have calculated the gradient at -2/3. Then I am looking to find the midpoints, of the line, which are [4/2, -2/2]. I have then calculated the perpendicular of the line AB, which had a gradient - 2/3. The perdendicular gradient is 2/3.
No, it isn't. The gradient of the perpendicular line is 3/2. If two perpendicular lines have slopes m1 and m2, then m1m2= -1.
 
HallsofIvy said:
No, it isn't. The gradient of the perpendicular line is 3/2. If two perpendicular lines have slopes m1 and m2, then m1m2= -1.

Sorry that is a typo error on my behalf now I have looked back.:o

The midpoints are; [4/2, - 2/2 ], [-1/2, 3/2 ]

With reference to the parametric equation I give the solution and not the original parametric equation.

So y = (3x - 8) / (2) = (3(4/2) - 8)) / (2) = -2/2

Kind regards

Casio(Thinking)
 

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