How can I find a point that is equidistant from two non-parallel lines?

[Physicsissuef]

Homework Statement

On the straight line x+y-8=0 find point on equal distance of the point M(2,8) and equal distance of x-3y+2=0

Homework Equations

$$d=\frac{|Ax + By + C|}{|\sqrt{A^2+B^2}|}$$

The Attempt at a Solution

Should I first find d (distance) from M(2,8) and x+y-8=0?

 

[sutupidmath]

Homework Statement

On the straight line x+y-8=0 find point on equal distance of the point M(2,8) and equal distance of x-3y+2=0

Homework Equations

$$d=\frac{|Ax + By + C|}{|\sqrt{A^2+B^2}|}$$

The Attempt at a Solution

Should I first find d (distance) from M(2,8) and x+y-8=0?

ok, here it is what i think you need to do: f


the distance among any two points is given with the equation

$$d=\sqrt{(x-x_1)^{2}+(y-y_1)^{2}}$$

Let $$M'(x',y')$$ be a point in the first line, and $$M''(x'',y'')$$ a point on the second line so from here we get:


$$\sqrt{(2-x'')^{2}+(8-y'')^{2}}= d=\sqrt{(2-x')^{2}+(8-y')^{2}}$$

Also use the fact that M', and M'' satisfy the equations of the lines, and see if you can come up with sth.

 

[Physicsissuef]

How to do that?

 

[HallsofIvy]

What exactly is the question? To find a point that is equi-distant from the line x- 3y+ 2= 0 and the point (2, 8)? There are an infinite number of such points: it is well known that the trace of points equidistant from a given point and a given line is a parabola having the given point as focus and the given line as directrix.

Probably the simplest way to find a single such point is to find the midpoint on the line through (2, 8) perpendicular to x- 3y+ 2= 0.

To find an equation for all such points (the equation of the parabola) set the distance from (x,y) to the line equal to the distance from (x,y) to the point equal:
$$\frac{x- 3y+ 2}{\sqrt{10}}= \sqrt{(x-2)^2+ (y-8)^2}$$
square both sides and simplify

 

[Physicsissuef]

Yes, you're right. I also realized that. Also we don't know if the distance is normal, so we don't know actually if the formula $$d=\frac{|Ax + By + C|}{|\sqrt{A^2+B^2}|}$$ does imply... I found one point M(3,5)

 

[HallsofIvy]

"We" don't? Distance from a point to a line (in fact, distance from anything to a line) is always the "shortest" distance and so is always measured perpendicular to the line.

Oh wait, for the first time I notice the part "On the straight line x+y-8=0"! There are either one or two points on that line that cross the parabola of points equidistant to the line and point. Your point (3, 5) satifies that x+ y- 8= 0.
Now, let's see- the distance from (3, 5) to the line x- 3y+ 2= 0 is, according to your formula
$$\frac{|3- 3(5)- 9|}{\sqrt{2}}= \frac{3}{\sqrt{2}}$$
The distance from (3, 5) to (2, 8) is $\sqrt{(3-2)^2+ (5-8)^2}= \sqrt{1+ 9}= \sqrt{10}$. Those don't see to be the same to me.

 

[Physicsissuef]

you're wrong.
It should be
$$\frac{|3- 3(5)+2|}{\sqrt{10}}= \sqrt{10}$$

 

[HallsofIvy]

Ooops!

 

[sutupidmath]

How to do that?

Well, when i look a lill bit closer to this problem, i think that it becomes much simpler this way, since you are required to find any point, regardless of what that is. There might be others also that fullfill that condition, but defenitely this will be one.
So we have two lines

$$L_1:x+y-8=0, \ \ \ \ and \ \ \ \ L_2:x-3y+2=0$$

Now: $$L_1:y=-x+8, \ \\ \ and \ \ \ L_2:y=\frac{1}{3}x+\frac{2}{3}$$

Since we are on the plane, and also since these two lines obviously are not parallel, then they must intersect each other, and this way have a common point, call it K(x',y'). Indeed, this point will be on two lines, and also will be of equal distance from any point and also from your point M(2,8).

So it is sufficient to find the point where these two lines intersect. By solving these eq. we get

$$4y-10=0=>y=\frac{5}{2}$$ and $$x=8-y=>x=8-\frac{5}{2}=\frac{11}{2}$$

So, $$K(\frac{11}{2},\frac{5}{2})\right)$$ is a point in two lines that is of equal distance from your point.

Well, we have to observe one more thing here, which actually makes all this be true, that M(2,8) does not lie in any of those two lines.