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Using Similar Triangles to solve for the equation of a line
Determine y-coordinate of a 3rd point from 2 given points and an x-coordinate.In Taxicab Geometry, what is the solution to d(P, A) = 2 d(P, B) for two points, A and B?Proving Similar triangles using SSSHow to estimate orientation errors of an image with respect to known data (line features)Determine y-coordinate of a 3rd point from 2 given points and an x-coordinate.Using similar triangles, find l?Find the relative radial position of a point within an ellipsoidHigh school geometry proof helpCurve for similar trianglesGeometry: move of “x” centimeter on a curveUsing the pythagorean theorem and similar triangles
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Consider points A=(−10,−4) and C=(8,5). The point B is on the line passing through A and C. The x-coordinate of B is −1. Determine the y-coordinate of the point B.
This question has been asked before. I am very much interested in a visual representation of solving this. It is part of a coursera self-paced no cert calculus course.
I have a very hard time equating the following solution pictorially with the use of similar triangles.
BD/DA = CE/EA where tow letter pairs are line segments.
DA= x-coordinate of D minus x-coordinate of A= −1+10=9
CE= y-coordinate of C minus y-coordinate of E= |5+4|=9
EA= x-coordinate of E minus x-coordinate of A= 8+10=18
so DB/9=918
Therefore, DB =9/2
Since B is above D, the y-coordinate of B is the y-coordinate of D plus the length of DB.
Thus the y-coordinate of B=−4+92=12
I am also fuzzy on Since B is above D, the y-coordinate of B is the y-coordinate of D plus the length of DB. What it was above.
Thanks,
Chris
geometry
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add a comment |
$begingroup$
Consider points A=(−10,−4) and C=(8,5). The point B is on the line passing through A and C. The x-coordinate of B is −1. Determine the y-coordinate of the point B.
This question has been asked before. I am very much interested in a visual representation of solving this. It is part of a coursera self-paced no cert calculus course.
I have a very hard time equating the following solution pictorially with the use of similar triangles.
BD/DA = CE/EA where tow letter pairs are line segments.
DA= x-coordinate of D minus x-coordinate of A= −1+10=9
CE= y-coordinate of C minus y-coordinate of E= |5+4|=9
EA= x-coordinate of E minus x-coordinate of A= 8+10=18
so DB/9=918
Therefore, DB =9/2
Since B is above D, the y-coordinate of B is the y-coordinate of D plus the length of DB.
Thus the y-coordinate of B=−4+92=12
I am also fuzzy on Since B is above D, the y-coordinate of B is the y-coordinate of D plus the length of DB. What it was above.
Thanks,
Chris
geometry
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$begingroup$
problem is solved at math.stackexchange.com/questions/919555/…
$endgroup$
– Chris Rigano
Jan 17 '15 at 5:33
add a comment |
$begingroup$
Consider points A=(−10,−4) and C=(8,5). The point B is on the line passing through A and C. The x-coordinate of B is −1. Determine the y-coordinate of the point B.
This question has been asked before. I am very much interested in a visual representation of solving this. It is part of a coursera self-paced no cert calculus course.
I have a very hard time equating the following solution pictorially with the use of similar triangles.
BD/DA = CE/EA where tow letter pairs are line segments.
DA= x-coordinate of D minus x-coordinate of A= −1+10=9
CE= y-coordinate of C minus y-coordinate of E= |5+4|=9
EA= x-coordinate of E minus x-coordinate of A= 8+10=18
so DB/9=918
Therefore, DB =9/2
Since B is above D, the y-coordinate of B is the y-coordinate of D plus the length of DB.
Thus the y-coordinate of B=−4+92=12
I am also fuzzy on Since B is above D, the y-coordinate of B is the y-coordinate of D plus the length of DB. What it was above.
Thanks,
Chris
geometry
$endgroup$
Consider points A=(−10,−4) and C=(8,5). The point B is on the line passing through A and C. The x-coordinate of B is −1. Determine the y-coordinate of the point B.
This question has been asked before. I am very much interested in a visual representation of solving this. It is part of a coursera self-paced no cert calculus course.
I have a very hard time equating the following solution pictorially with the use of similar triangles.
BD/DA = CE/EA where tow letter pairs are line segments.
DA= x-coordinate of D minus x-coordinate of A= −1+10=9
CE= y-coordinate of C minus y-coordinate of E= |5+4|=9
EA= x-coordinate of E minus x-coordinate of A= 8+10=18
so DB/9=918
Therefore, DB =9/2
Since B is above D, the y-coordinate of B is the y-coordinate of D plus the length of DB.
Thus the y-coordinate of B=−4+92=12
I am also fuzzy on Since B is above D, the y-coordinate of B is the y-coordinate of D plus the length of DB. What it was above.
Thanks,
Chris
geometry
geometry
asked Jan 17 '15 at 5:15
Chris RiganoChris Rigano
1081
1081
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problem is solved at math.stackexchange.com/questions/919555/…
$endgroup$
– Chris Rigano
Jan 17 '15 at 5:33
add a comment |
$begingroup$
problem is solved at math.stackexchange.com/questions/919555/…
$endgroup$
– Chris Rigano
Jan 17 '15 at 5:33
$begingroup$
problem is solved at math.stackexchange.com/questions/919555/…
$endgroup$
– Chris Rigano
Jan 17 '15 at 5:33
$begingroup$
problem is solved at math.stackexchange.com/questions/919555/…
$endgroup$
– Chris Rigano
Jan 17 '15 at 5:33
add a comment |
2 Answers
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One thing that seems apparent is that since B lies right at the average of the x values of the triangles, then the triangles should be congruent. Therefore the y value of B should be the average of A and C y values. $frac12$
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add a comment |
$begingroup$
The last sentence of the fifth paragraph should have read,
Thus the $y$-coordinate of B is $−4+dfrac92=dfrac12$.
which is the correct answer.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
One thing that seems apparent is that since B lies right at the average of the x values of the triangles, then the triangles should be congruent. Therefore the y value of B should be the average of A and C y values. $frac12$
$endgroup$
add a comment |
$begingroup$
One thing that seems apparent is that since B lies right at the average of the x values of the triangles, then the triangles should be congruent. Therefore the y value of B should be the average of A and C y values. $frac12$
$endgroup$
add a comment |
$begingroup$
One thing that seems apparent is that since B lies right at the average of the x values of the triangles, then the triangles should be congruent. Therefore the y value of B should be the average of A and C y values. $frac12$
$endgroup$
One thing that seems apparent is that since B lies right at the average of the x values of the triangles, then the triangles should be congruent. Therefore the y value of B should be the average of A and C y values. $frac12$
answered Jan 17 '15 at 5:35
turkeyhundtturkeyhundt
6,85511025
6,85511025
add a comment |
add a comment |
$begingroup$
The last sentence of the fifth paragraph should have read,
Thus the $y$-coordinate of B is $−4+dfrac92=dfrac12$.
which is the correct answer.
$endgroup$
add a comment |
$begingroup$
The last sentence of the fifth paragraph should have read,
Thus the $y$-coordinate of B is $−4+dfrac92=dfrac12$.
which is the correct answer.
$endgroup$
add a comment |
$begingroup$
The last sentence of the fifth paragraph should have read,
Thus the $y$-coordinate of B is $−4+dfrac92=dfrac12$.
which is the correct answer.
$endgroup$
The last sentence of the fifth paragraph should have read,
Thus the $y$-coordinate of B is $−4+dfrac92=dfrac12$.
which is the correct answer.
answered Jan 17 '15 at 6:03
Mark LaoMark Lao
531210
531210
add a comment |
add a comment |
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problem is solved at math.stackexchange.com/questions/919555/…
$endgroup$
– Chris Rigano
Jan 17 '15 at 5:33