Perpendicular Distance Question

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January 4th 2013, 11:00 AM
HelenMc9

Perpendicular Distance Question

How would I do this?

Find the equation of the lines that pass through the point (-3, -4) and that are a distance of √10 from the point (2,1).

I believe I need to use this formula to solve the question:
|ax1 + by1 + c| / √(a² + b²)

Probably isn't hard but any help would be greatly appreciated!
January 4th 2013, 11:20 AM
MarkFL

Re: Perpendicular Distance Question

I would approach this problem by finding the lines through (-3,-4) and tangent to the circle:

$(x-2)^2+(y-1)^2=10$

Using the point-slope formula, let the tangent lines have the form:

$y+4=m(x+3)$

$y=m(x+3)-4$

Now, substitute into the circle for y to get a quadratic in x, and set the discriminant to zero (do you know why?) to get a quadratic in m.
January 4th 2013, 11:30 AM
HelenMc9

Re: Perpendicular Distance Question

Thanks a million for the help but I'm totally confused by that sorry, I don't really get any of that! I'm doing some Co-ordinate Geometry homework and the topic is 'The Perpendicular Distance from a point to a line' and we haven't come across anything to do with tangents or circles so I think we're expected to do it that way. Is there any other way you can think of to do it? :)
January 4th 2013, 11:36 AM
vjdm

Re: Perpendicular Distance Question

firstly find the eqation of a line passing thru (-3,-4) by point slope form then apply formula of a dist b/w pont and a line to find m after that put the value of m to get the required equation
January 4th 2013, 11:42 AM
vjdm

Re: Perpendicular Distance Question

point slope form
y-y1=m(x-x1)
y+4=m(x+3)
i.e. mx-y+3m-4=0--------- I
then apply perpendicular dist b/w point and a line
|ax1 + by1 + c| / √(a² + b²) by equating tht with √10
find m from here then put tht in I
January 4th 2013, 11:59 AM
MarkFL

Re: Perpendicular Distance Question

Quote:

Originally Posted by HelenMc9

Thanks a million for the help but I'm totally confused by that sorry, I don't really get any of that! I'm doing some Co-ordinate Geometry homework and the topic is 'The Perpendicular Distance from a point to a line' and we haven't come across anything to do with tangents or circles so I think we're expected to do it that way. Is there any other way you can think of to do it? :)

Sorry, I was just musing about how I might approach this problem from a different angle.

As I showed in my first post, the tangent lines will have the form:

$y+4=m(x+3)$

So, let's put that into standard form:

$-mx+y+(4-3m)=0$

So, for the standard form of a line we identify:

$A=-m,\,B=1,\,C=4-3m$

Now, our formula for the distance d between a point $(x_1,y_1)$ and the line $Ax+By+C=0$ is:

$d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+C^2}}$

We are given:

$d=\sqrt{10},\,(x_1,y_1)=(2,1)$ and so we have:

$\sqrt{10}=\frac{|-2m+1+4-3m|}{\sqrt{(-m)^2+(1)^2}}$

$\sqrt{10}=\frac{5|m-1|}{\sqrt{m^2+1}}$

Now, square the equation:

$10=\frac{25(m-1)^2}{m^2+1}$

Now, solve for m. You will get two values, corresponding to the two possible lines.

What do you find?
January 4th 2013, 12:35 PM
HelenMc9

Re: Perpendicular Distance Question

10m² + 10 = 25(m²-2m+1)
10m² + 10 = 25m² -50m + 25
15m²-50m+15=0
3m²-10m+3=0
3m²-9m-m+3=0
3m(m-3)-1(m-3) = 0
(3m-1)(m-3) = 0
3m=1 or m=3

so m=1/3 or m=3. Is that right?

And then I suppose I can fill in the two separate values for m into -mx + y + (4-3m) = 0 to get the two separate equations?
January 4th 2013, 12:39 PM
MarkFL

Re: Perpendicular Distance Question

Yes, good work! (Clapping)(Rock)(Cool)
January 4th 2013, 12:50 PM
HelenMc9

Re: Perpendicular Distance Question

Thank-you very much! I understand it now :)
January 4th 2013, 12:53 PM
MarkFL

Re: Perpendicular Distance Question

Glad to help, particularly for someone who is a math helper's dream...you stated the problem clearly in its entirety with your thoughts on what you needed to use, then responded to feedback with work too. That is rare. :D