Thread: 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!

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



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





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.

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?

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

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

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:



So, let's put that into standard form:



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



Now, our formula for the distance d between a point and the line is:



We are given:

and so we have:





Now, square the equation:



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

What do you find?

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?

Yes, good work!

Thank-you very much! I understand it now

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.