MHB Finding the Equation of a Line

  • Thread starter Thread starter Casio1
  • Start date Start date
  • Tags Tags
    Line
Click For Summary
The discussion revolves around finding the equation of a line related to a circle problem involving points A(5, -3) and B(-1, 1). The gradient of line AB is calculated as -2/3, leading to a midpoint of [4/2, -2/2] and a perpendicular gradient of 3/2. The derived equation for the perpendicular bisector is y = 2/3(x - 6/2), but there is confusion regarding its equivalence to the parametric equation y = (3x - 8) / 2, as the results do not match. Clarifications reveal that the correct gradient for the perpendicular line should be 3/2, highlighting a misunderstanding in the calculations. The conversation emphasizes the importance of correctly applying the relationships between slopes and equations in line geometry.
Casio1
Messages
86
Reaction score
0
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:
 
Mathematics news on Phys.org
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)
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
View full post »

Similar threads

MHB [ASK] A Line Intercepting A Circle
  • · Replies 4 ·
Replies
4
Views
1K
B Solving the Heart Curve Equation: 'y =
  • · Replies 18 ·
Replies
18
Views
4K
MHB Math Cumulative Review: Find the equation of a Line
  • · Replies 2 ·
Replies
2
Views
2K
B Arithmetic mean of an infinite number of points?
  • · Replies 20 ·
Replies
20
Views
2K
B Comparing Gradient: Highest & Lowest, Vector/Scalar?
  • · Replies 3 ·
Replies
3
Views
1K
MHB Finding the Equation of a Line Given a Point and Slope
  • · Replies 2 ·
Replies
2
Views
1K
MHB Parametric Eqs: Find Line & Plane, Find Triangle Area
  • · Replies 4 ·
Replies
4
Views
1K
MHB Quadratic equations ( solving for zeros/roots) trickey
  • · Replies 1 ·
Replies
1
Views
1K
MHB System of Equations: Find Triples $(x,y,z)$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Fixed point,, Jacobi- & Newton Method, Linear Systems
  • · Replies 7 ·
Replies
7
Views
2K