Distance from a point to line problem in need of some help

  • Thread starter Wats
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  • #1
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Distance from a point to line problem....in need of some help

Homework Statement


Find an equation of the line bisecting the angle from the first line to the second.

24x-7y+1=0, 3x+4y-5=0


Homework Equations


|Ax+By+C|/√A^2 + B^2


The Attempt at a Solution


|24x+7y+1|/√(625) = |3x+4y-5|/√(25)

The answer the book gave was 39x+13y-24. These problems are going right over my head. I think it's the position of the point and the absolute values along with the square roots that are throwing me off. I'd really appreciate any help.
Thanks in advance.
 

Answers and Replies

  • #2
HallsofIvy
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You seem to be completely confused as to what you are doing. You titled this "Distance from a point to line" and the "relevant equation" you give is for the distance from the point (x, y) to the line Ax+ By+ C= 0.

But the problem posed, "Find an equation of the line bisecting the angle from the first line to the second" has nothing to do with "distance" not is there any point.

Please check exactly what the problem says.
 
  • #3
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That is exactly what the problem says. Those were the instructions given and the two equations given. The title of the section in this chapter is called "Distance from a point to a line". The way I understand it is there is a point that is equidistant from each line given, and you're suppose to find the equation for that particular point. That is why I set each equation equal to each other, but I can't seem to get past that.
 
  • #4
HallsofIvy
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Ah. That isn't the way I would have done the problem but perhaps this is easier. If two lines intersect then every point on the lines (there are two) that bisect the the angles they make are equidistant from the two lines.

Let (x, y) be a point on the bisector. Write out the formula for the distance from (x, y) to each of the two lines and set them equal. Solve for y as a function of x. That will be the equation of the bisector.
 

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