Intersection of Two Lines

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    Intersection Lines
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Discussion Overview

The discussion revolves around finding the intersection point of a normal line through the point (3,-4) to the line defined by the equation 10x+4y-101=0. Participants explore various methods, including vector approaches and slope calculations, to solve the problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the dot product of the vector from (3,-4) to a point (x,y) with the vector (-4/10,-10/4) to find the intersection, but another participant questions the direction of this vector.
  • Another participant correctly identifies the slope of the line as -10/4, leading to the conclusion that the normal line has a slope of 2/5, and provides the equation for the normal line.
  • A participant expresses confusion about the derivation of the normal line's equation and proposes an alternative vector approach using (x-3,y+4) and (2,5), leading to a different intersection point.
  • Some participants emphasize the use of vectors for the solution, while others suggest a simpler method based on slope and point-slope form.
  • There is a discussion about the concept of direction vectors, with one participant noting that there are infinitely many vectors pointing in the same direction and clarifying the correct use of the dot product for perpendicular lines.

Areas of Agreement / Disagreement

Participants exhibit disagreement on the preferred method of solution, with some advocating for vector approaches while others prefer traditional slope-based methods. The discussion remains unresolved regarding the most effective technique to find the intersection point.

Contextual Notes

Participants express uncertainty about the correct application of vector methods and the implications of using different direction vectors. There are also unresolved mathematical steps in the derivation of the intersection point.

Who May Find This Useful

This discussion may be useful for students and enthusiasts of geometry and vector mathematics, particularly those interested in the intersection of lines and the application of different mathematical approaches to solve problems.

thomasrules
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I have tried but doesn't work out well...

Find the point at which the normal through the point (3,-4) to the line 10x+4y-101=0 intersects the line.

Originaly I thought that it should be found by doing the dot product of:

(x-3,y+4)dot(-4/10,-10/4)= 0
 
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thomasrules said:
I have tried but doesn't work out well...
Find the point at which the normal through the point (3,-4) to the line 10x+4y-101=0 intersects the line.
Originaly I thought that it should be found by doing the dot product of:
(x-3,y+4)dot(-4/10,-10/4)= 0
I can't help but wonder what your reasoning is! (The vector (-4/10, -10/4) is not in the direction of that line. The vector (4, -10) or (-4, 10) are.)

The line 10x+ 4y- 101= 0 can be written as y= (-10/4)x+ 101/10. That has slope -10/4 so a normal line has slope 4/10= 2/5.
The line through (3,-4) with that slope can be written y= (2/5)(x-3)- 4 or, without fractions, 4y= 2x- 26

Solve the equations 10x+ 4y- 101= 0 and 4y= 2x- 26 for x and y.
 
HallsofIvy said:
The line through (3,-4) with that slope can be written y= (2/5)(x-3)- 4 or, without fractions, 4y= 2x- 26
Solve the equations 10x+ 4y- 101= 0 and 4y= 2x- 26 for x and y.

ok i don't get how you get y= (2/5)(x-3)- 4

can't you do (x-3,y+4)dot(2,5)= 0 and then substitute

can't I also write it as r=(3,-4)+t(2,5)?

LOL i tried doing it my way and i got as an answer to be (187/14,-171/21)
 
Last edited:
lol can someone help me further :(
 
Are you determined to do this using vectors? Since this problem is entirely in 2 dimensions, I see no reason not to do it in the simplest possible way. The line 10x+4y-101=0 has slope -5/2. (Solve for y:
4y= -10x+ 101 so y= (-10/4)x+ 101/4 and -10/4= -5/2. Any line perpendicular to that must have slope 2/5.

The "point slope" form for a line, with slope m going through (a, b), is
y= m(x-a)+ b. That clearly has slope m and, just as clearly, when x= a, y= b.
 
yea but I'm doing geometry and you was is like going back to grade 10...We learned vectors and probably have to do it taht way
 
yea i''m really struggling with this one...

I understand that if you change 10x+4y-101=0 to y=(101-10x)\4 that the direction vector for that line would be (4,-10) am I right?

then since they are perpendicular I would take the point (3,-4) and subract it to make the direction vector for the other line...so

(x-3,x+4)dot(4,-10)=0 AM I RIGHT?

P.S. I have to do it this way because I have to use vectors
 
Part of your problem is that you keep saying the direction vector. There are an infinite number of vectors pointing in the direction of a given line, all of different lengths.

Yes, if x increases by 4 (say from 0 to 4), y decreases by 10 (from 101/4 to 61/4) so a vector pointing along the line is (4, -10). You could just as easily use (2, -5) or (-2, 5).

Yes, the equation of the perpendicular line can be written
(x- 3, y+ 4)dot(4, -10)= 0 (You have "(x- 3, x+ 4)" but I assume that is a typo. If you multiply that out, you get 4(x- 3)- 10(y+ 4)= 4x- 12- 10y- 40= 4x- 10y- 52=0 which is equivalent to the equation I gave.

Solve the two equations, 4x- 10y- 52=0 and 10x+4y-101=0 simultaneously to find the point at which they intersect.
 

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