MHB Finding the Equation of a Line

[Casio1]

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

b_AB : 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

 

[Ackbach]

Could you please state the original problem verbatim?

 

[Casio1]

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

 

[Sudharaka]

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.

 

[HallsofIvy]

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.

 

[Casio1]

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.

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)