Minimum Distance

[TangoFoxtrotGolf]

Find all values of b such that the minimum distance from the point (2,0) to the line Y = (4/3)X + b is 5.

d = 5 = ((Y - 0)[sup:u4ekfm1r]2[/sup:u4ekfm1r] + (X - 2)[sup:u4ekfm1r]2[/sup:u4ekfm1r])[sup:u4ekfm1r]1/2[/sup:u4ekfm1r]

Squaring both sides,

25 = (Y - 0)[sup:u4ekfm1r]2[/sup:u4ekfm1r] + (X - 2)[sup:u4ekfm1r]2[/sup:u4ekfm1r]
25 = (Y)[sup:u4ekfm1r]2[/sup:u4ekfm1r] + (X[sup:u4ekfm1r]2[/sup:u4ekfm1r] - 4X + 4)
25 = Y[sup:u4ekfm1r]2[/sup:u4ekfm1r] + X[sup:u4ekfm1r]2[/sup:u4ekfm1r] - 4X + 4

Solve for Y[sup:u4ekfm1r]2[/sup:u4ekfm1r], but purposely leave Y[sup:u4ekfm1r]2[/sup:u4ekfm1r] negative

-Y[sup:u4ekfm1r]2[/sup:u4ekfm1r] = X[sup:u4ekfm1r]2[/sup:u4ekfm1r] - 4X - 21

Take the equation for the line and square it,

Y[sup:u4ekfm1r]2[/sup:u4ekfm1r] = ((4/3)X + b)[sup:u4ekfm1r]2[/sup:u4ekfm1r]
Y[sup:u4ekfm1r]2[/sup:u4ekfm1r] = (16/9)X[sup:u4ekfm1r]2[/sup:u4ekfm1r] + (8/3)bX + b[sup:u4ekfm1r]2[/sup:u4ekfm1r]

Now add the other equation to this one,

Y[sup:u4ekfm1r]2[/sup:u4ekfm1r] = (16/9)X[sup:u4ekfm1r]2[/sup:u4ekfm1r] + (8/3)bX + b[sup:u4ekfm1r]2[/sup:u4ekfm1r]
-Y[sup:u4ekfm1r]2[/sup:u4ekfm1r] = X[sup:u4ekfm1r]2[/sup:u4ekfm1r] - 4X - 21
------------------------------------------------------------------------
0 = (25/9)X[sup:u4ekfm1r]2[/sup:u4ekfm1r] + (8/3)bX - 4X - 21 + b[sup:u4ekfm1r]2[/sup:u4ekfm1r]

But, now what. If I try to isolate b on the left side, I also move X to the left side. If I then divide both sides by X to get rid of X in the (8/3)bX term, I also leave X in the b[sup:u4ekfm1r]2[/sup:u4ekfm1r] term.

 

[DrMike]

The key here is that the minimum distance to the line is 5. Therefore your point (X,Y) is not just any old point on the line, but a special one - the line from (2,0) to (X,Y) is at right angles to Y = (4/3)X + b.

This gives you an extra equation, you'll be able to eliminate X.

Let us know if you're still stuck...

 

[TangoFoxtrotGolf]

Dr. Mike,

It took me a while to get back to this. I mostly only get to work on this during the week end and holidays.

Finally, got the right answer, by
Finding the equation of the perpendicular line,
Setting the X sides of both equations equal to each other to solve for X
Using that X value in either equation to find Y.
Using the distance equation and the two Points to solve for b.

There ends up being two answers for b since there are two parallel lines that are 5 units distant from (2,0).

Thanks for the help.

 

[Mrspi]

You seem to have "gotten" it...

You might want to look at this website, as well, for a general formula for the shortest distance from a point to a line:

http://www.mathwords.com/d/distance_point_to_line.htm