Parametric equations

[Probability]

I have x = t + 2, and 2y = 3t - 2. I wish to eliminate t. This is what I have tried.

t = x - 2,

2y = 3t - 2 = 3(x - 2) - 2 = 3x - 6 - 2 =

y = (3x - 8) / (2). I am unsure about this although it may look good on paper, the reason;

Take my points A(5, - 3), B(- 1, 1), and C(0, 2)

My gradient is 2/3

the midpoints are; [4/2, 2/2], and [- 1/2, 2/2]

My equation of the perpendicular bisector AB is

y - y₁ = m(x - x₁)

y - ²/₂ = ²/₃(x - 5)

y = ²/₃x - ¹⁰/₃ - ²/₂

y = ²/₃x - ⁷/₃

Ok both the above should give the same answer from the same input data.

y = (3x - 8) / (2) = (3(5) - 8)) / (2) = 3.5. (I have no coordinates of 3.5 so I am assuming I go this wrong)?

OK the equation of the line.

y = ²/₃x - ⁷/₃

Enter same data.

y = ²/₃(5) - ⁷/₃ = 1. (OK I have a y value of 1 at B, so if x₁ = 5 and y₂ = 1 this seems to suggest that I have found one coorodinate of the points from A = x_1, 5, and B = y₂ 1.

If I now plug into the equation B = x₂ - 1, then I get the result A = y₁ = - 3.

So it seems that my equation will find the coordinates of the points A and B, but my parametric equation above will not?

Can anyone spot anything obvious?

 

[Deleted member 4993]

I have x = t + 2, and 2y = 3t - 2. I wish to eliminate t. This is what I have tried.

t = x - 2,

2y = 3t - 2 = 3(x - 2) - 2 = 3x - 6 - 2 =

y = (3x - 8) / (2). I am unsure about this although it may look good on paper, the reason;

Take my points A(5, - 3), B(- 1, 1), and C(0, 2) A is not on the line whose equation you have derived. Neither is C or B.

My gradient is 2/3

the midpoints are; [4/2, 2/2], and [- 1/2, 2/2]

My equation of the perpendicular bisector AB is

y - y₁ = m(x - x₁)

y - ²/₂ = ²/₃(x - 5)

y = ²/₃x - ¹⁰/₃ - ²/₂

y = ²/₃x - ⁷/₃

Ok both the above should give the same answer from the same input data.

y = (3x - 8) / (2) = (3(5) - 8)) / (2) = 3.5. (I have no coordinates of 3.5 so I am assuming I go this wrong)?

OK the equation of the line.

y = ²/₃x - ⁷/₃

Enter same data.

y = ²/₃(5) - ⁷/₃ = 1. (OK I have a y value of 1 at B, so if x₁ = 5 and y₂ = 1 this seems to suggest that I have found one coorodinate of the points from A = x_1, 5, and B = y₂ 1.

If I now plug into the equation B = x₂ - 1, then I get the result A = y₁ = - 3.

So it seems that my equation will find the coordinates of the points A and B, but my parametric equation above will not?

Can anyone spot anything obvious?

.

 

[HallsofIvy]

I have x = t + 2, and 2y = 3t - 2. I wish to eliminate t. This is what I have tried.

t = x - 2,

2y = 3t - 2 = 3(x - 2) - 2 = 3x - 6 - 2 =

y = (3x - 8) / (2). I am unsure about this although it may look good on paper, the reason;

Yes, that is competely correct. It not only "looks good on paper", it is good.

Take my points A(5, - 3), B(- 1, 1), and C(0, 2)

What? Where did these points come from? What do they have to do with the original line? They are certainly NOT on the line x= t+ 2, 2y= 3t- 2: if x= t+ 2= 5, then t= 3 so that 2y= 6- 2= 4 and y= 2, not -3. If x= t+ 2= -1, then t= -3 so that 2y= -9- 2= -11 and y= -11/2, not 1. Finally, if x= t+ 2= 0, t= -2 so that 2y= -6- 2= -8 and y= -4, not 2.

My gradient is 2/3

Gradient of what is 2/3? The gradient of the original line is 3/2, not 2/3. The gradient of AB is (5+1)/(-3-1)= 6/-4= -3/2, the gradient of AC is (5- 0)/(-3-2)= 5/-6= -5/6, and the gradient of BC is (-1- 0)/(1- 2)= -1/-1= 1. NONE of the possible lines have gradient "2/3".

the midpoints are; [4/2, 2/2], and [- 1/2, 2/2]

Midpoints of what? I can see that the midpoint of AB is ((5-1)/2,(-3+1)/2)= (2, -1), the midpoint of BC is ((0-1)/2, (2+ 1)/2)= (-1/1, 3/2), and the midpoint of AC is ((5+ 0)/2, (-3+ 2)/2)= (5/2, -1/2). NONE of the possible line segments have the midpoints you give.

My equation of the perpendicular bisector AB is

y - y₁ = m(x - x₁)

y - ²/₂ = ²/₃(x - 5)

y = ²/₃x - ¹⁰/₃ - ²/₂

y = ²/₃x - ⁷/₃

Ok both the above should give the same answer from the same input data.

y = (3x - 8) / (2) = (3(5) - 8)) / (2) = 3.5. (I have no coordinates of 3.5 so I am assuming I go this wrong)?

OK the equation of the line.

y = ²/₃x - ⁷/₃

Enter same data.

y = ²/₃(5) - ⁷/₃ = 1. (OK I have a y value of 1 at B, so if x₁ = 5 and y₂ = 1 this seems to suggest that I have found one coorodinate of the points from A = x_1, 5, and B = y₂ 1.

If I now plug into the equation B = x₂ - 1, then I get the result A = y₁ = - 3.

So it seems that my equation will find the coordinates of the points A and B, but my parametric equation above will not?

Can anyone spot anything obvious?

 

[Probability]

Yes, that is competely correct. It not only "looks good on paper", it is good.


What? Where did these points come from? What do they have to do with the original line? They are certainly NOT on the line x= t+ 2, 2y= 3t- 2: if x= t+ 2= 5, then t= 3 so that 2y= 6- 2= 4 and y= 2, not -3. If x= t+ 2= -1, then t= -3 so that 2y= -9- 2= -11 and y= -11/2, not 1. Finally, if x= t+ 2= 0, t= -2 so that 2y= -6- 2= -8 and y= -4, not 2.


Gradient of what is 2/3? The gradient of the original line is 3/2, not 2/3. The gradient of AB is (5+1)/(-3-1)= 6/-4= -3/2, the gradient of AC is (5- 0)/(-3-2)= 5/-6= -5/6, and the gradient of BC is (-1- 0)/(1- 2)= -1/-1= 1. NONE of the possible lines have gradient "2/3".


Midpoints of what? I can see that the midpoint of AB is ((5-1)/2,(-3+1)/2)= (2, -1), the midpoint of BC is ((0-1)/2, (2+ 1)/2)= (-1/1, 3/2), and the midpoint of AC is ((5+ 0)/2, (-3+ 2)/2)= (5/2, -1/2). NONE of the possible line segments have the midpoints you give.

I can only appologise for the above, I knew when writing it that combining parametric equations and my other points were a bad move. I will start again on a new thread.