Analytic Geometry World Problem

AI Thread Summary
To find the coordinates of point T, which lies on the line 2x+y=3 and is 13 units from point S (2, -1), a system of equations is used involving the distance formula. The method involves substituting y with -2x+3 in the distance equation, leading to a quadratic equation that yields two solutions. This approach is validated by multiple participants, confirming its correctness. Drawing a circle with a diameter of 13 units centered at S aids in visualizing the problem. Ultimately, the discussion emphasizes the validity of the method and the expected two solutions for point T.
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Homework Statement


Points S and T lie on the line 2x+y=3. If the length ST is 13 units, and the coordinates of S are (2, -1), determine all possible coordinate pairs for T, correct to two decimal places.

For simplicity, determine the coordinates for T, relative to S from the left.

Homework Equations


a^2=b^2+c^2 (in this problem, the distance formula.)

The Attempt at a Solution


I used a system of equations:

y=-2x+3

(2-x)^2+(-1-y)^2=13^2

This system is quite tedious. I used WolframAlpha to compute the solutions, however, they did not match the answer in my textbook.

Is this approach correct?
 
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S.R said:

Homework Statement


Points S and T lie on the line 2x+y=3. If the length ST is 13 units, and the coordinates of S are (2, -1), determine all possible coordinate pairs for T, correct to two decimal places.

For simplicity, determine the coordinates for T, relative to S from the left.


Homework Equations


a^2=b^2+c^2 (in this problem, the distance formula.)


The Attempt at a Solution


I used a system of equations:

y=-2x+3

(2-x)^2+(-1-y)^2=13^2

This system is quite tedious. I used WolframAlpha to compute the solutions, however, they did not match the answer in my textbook.

Is this approach correct?

Think of a circle with a diameter of 13 at point (2,-1). It helps if you actually draw this. To get to a coordinate or point on the arc you would...
 
Kingstrick, that is exactly what he is doing. S.R., yes, that is a valid method. of course, you can replace the y in the circle equation by -2x+ 3 so that it becomes
(2- x)^2+ (-1-(-2x+3))^2= (2- x)^2+ (2x- 4)^2= 13. Multiplying that out will give you a quadratic equation which will have 2 solutions because, of course, this line is a diameter of the circle and so crosses it twice.
 
HallsofIvy said:
Kingstrick, that is exactly what he is doing. S.R., yes, that is a valid method. of course, you can replace the y in the circle equation by -2x+ 3 so that it becomes
(2- x)^2+ (-1-(-2x+3))^2= (2- x)^2+ (2x- 4)^2= 13. Multiplying that out will give you a quadratic equation which will have 2 solutions because, of course, this line is a diameter of the circle and so crosses it twice.
<br /> <br /> Thanks HallsofIvy!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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