Advertisement
Help Keep Boards Alive. Support us by going ad free today. See here: https://subscriptions.boards.ie/.
If we do not hit our goal we will be forced to close the site.

Current status: https://keepboardsalive.com/

Annual subs are best for most impact. If you are still undecided on going Ad Free - you can also donate using the Paypal Donate option. All contribution helps. Thank you.
https://www.boards.ie/group/1878-subscribers-forum

Private Group for paid up members of Boards.ie. Join the club.

The line - higher math problem

  • 11-05-2011 05:54PM
    #1
    Closed Accounts Posts: 147 ✭✭


    the lenght of the perpendicular to a line from the origin is 5 units . the line passes through the point (3,5) . Find the equations of two such lines.


    how to ?


Comments

  • Registered Users, Registered Users 2 Posts: 47 ShonyBoulders


    Well if it's a line from the origin...


  • Registered Users, Registered Users 2 Posts: 927 ✭✭✭Maybe_Memories


    Areq wrote: »
    the lenght of the perpendicular to a line from the origin is 5 units . the line passes through the point (3,5) . Find the equations of two such lines.


    how to ?

    Throw the info into your distance formula, get the slope of the line and the equation of the line, then use all that to get the perpendicular lines. you'll probably get x^2 and y^2 so that'll give you the two possibilities


  • Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    Use the general form of the line: y = mx +c, and try to use the given conditions to find m and c, as follows:

    Sub. in the point (3, 5), and use that to write c in terms of m, so now you have the equation of the line in terms of m alone (and x and y, obviously).

    Then, use the formula for the distance from a point to a line to write a second equation. This should be in terms of m alone, and be of the form of a modulus divided a square root equal to a constant.

    Multiple across by the denominator, and square both sides and you'll have a quadratic in m. Should be ok from there.


Advertisement