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

In summary, to determine the point on the line 5x-4y-2=0 where the normal that passes through the point (7,-2) intersects the line, you must first find the slope of the normal by taking the negative reciprocal of the slope of the given line. Then, using the slope and the given point, you can find the equation of the normal using the point-slope form. Finally, solve for the point of intersection by setting the equations of the normal and the given line equal to each other and solving for the coordinates.
  • #1
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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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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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. :(
 
  • #6
After some calculations I've come to the possible answer of (-2,-3).

Is this correct?
 
  • #7
Scratch that, I got (2,2)
 
  • #8
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.
 

1. What is a normal line?

A normal line is a line that is perpendicular to another line, and it intersects the other line at a 90-degree angle.

2. How do you determine the point on a line where a normal passes through a point?

The point on a line where a normal passes through a point can be determined by finding the perpendicular distance from the given point to the line. This can be done by using the formula d = |ax0 + by0 + c| / √(a^2 + b^2), where (x0, y0) is the given point and ax + by + c = 0 is the equation of the line.

3. What is the purpose of finding the point where a normal intersects a line?

Finding the point where a normal intersects a line is useful in many applications such as geometry, physics, and engineering. It can help determine the shortest distance between a point and a line, or the angle between two lines.

4. Can a normal line intersect a line at more than one point?

No, a normal line can only intersect a line at one point. This is because a normal line is perpendicular to the line it intersects, and two lines that are perpendicular can only intersect at one point.

5. How can the normal line concept be applied in real life situations?

The concept of normal lines can be applied in various real-life situations, such as finding the shortest distance between a point and a line (useful in navigation and construction), determining the angle of incidence and reflection in optics, and calculating the force of impact in physics.

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