Determine point on line where normal passes thr a point intersects the line.

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Homework Help Overview

The problem involves determining the point on the line defined by the equation 5x-4y-2=0 where a normal line passing through the point (7,-2) intersects this line. The subject area pertains to geometry and algebra, specifically the concepts of lines, slopes, and normal lines.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the slope of the given line and the corresponding slope of the normal line. There are attempts to derive the equation of the normal line through the point (7,-2) and questions about how to find the intersection with the original line. Some participants express confusion about the steps involved in solving for the intersection point.

Discussion Status

The discussion includes various attempts to derive equations and find intersections, with some participants expressing uncertainty about their progress. There is no explicit consensus on the correct answer, and multiple interpretations of the problem are being explored.

Contextual Notes

Some participants mention constraints related to missed lessons and express feelings of being lost in the problem-solving process. There are indications of varying levels of understanding among participants regarding the concepts involved.

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Homework Statement



Determine the point on the line 5x-4y-2=0 where the normal that passes through the point (7,-2) intersects the line.

The Attempt at a Solution



I'm completely and hopelessly lost. Any help or tips would be greatly appreciated. Thanks so much in advance!
 
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Presumably you know that the normal to y= mx+ b has slope -1/m.
5x- 4y- 2= 0 is the same as 4y= 5x- 2 or y= (5/4)x- 1/2. Its slope is 5/4 so the slope of any normal to it is -4/5.

Now, what is the equation of the line through (7, -2) with slope -4/5?
 
Now, what is the equation of the line through (7, -2) with slope -4/5?

Well a Vector Equation for that would be r = (7,-2) + t(-4,5) right? So where do I go from here?

Am I supposed to find the parametric equations of each and find the point of intersection? If so, what steps do I take from the ones mentioned above?
 
Since you are given one line in Cartesian, xy, form, it would be better to write the second line that way as well. I would be very surprised if you were working with vectors and parametric equations but did not know the "point, slope" form for a line in the plane.
 
So for the first line I have:

x = 7 - 4t
y= -2 + 5t

for the second line:

x = (4y + 2)/5

y = (5x - 2)/4


Now normally I'd equate the two, and solve. However there is no 't' value in the 2nd line from what I gathered? Am I missing something?

Sorry if I'm being a little frustrating, I had pneumonia and missed nearly all the lessons. :(
 
After some calculations I've come to the possible answer of (-2,-3).

Is this correct?
 
Scratch that, I got (2,2)
 
If you have a line with the equation of 5x-4y-2=0, you can figure out the slope of it's normal. If you have the slope and a point which the normal crosses (7; -2), all that you lack from normal equation y=kx+b is b. You can figure out slope from the other equation and a point with it's x and y coordinates is given. Sou you have system of linear equations to solve, where the solution is the crosspoint. I hope that helps.
 

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