# Finding the Equation of a Line

#### Casio

##### Member
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)

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

##### Indicium Physicus
Staff member
Could you please state the original problem verbatim?

#### Casio

##### Member
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

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
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.

#### Casio

##### Member
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 