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Quick question

  • 20-04-2010 8:15pm
    #1
    Registered Users, Registered Users 2 Posts: 237 ✭✭


    The teacher won't explain it again as I was absent, how do you do this question:

    L is the line 3x-y-11=0. The line K a(-3, 9) and b(6, r). K is perpendicular to L. Find R.


Comments

  • Closed Accounts Posts: 1,207 ✭✭✭Ally7


    Moromaster wrote: »
    The teacher won't explain it again as I was absent, how do you do this question:

    L is the line 3x-y-11=0. The line K a(-3, 9) and b(6, r). K is perpendicular to L. Find R.


    Well, you know the x value for that point is 6. R is obviously a y value so sub in 6 for x. What you get as the y value after you do out the equation is r.

    Hope this helps!


  • Registered Users, Registered Users 2 Posts: 237 ✭✭Moromaster


    Ally7 wrote: »
    Well, you know the x value for that point is 6. R is obviously a y value so sub in 6 for x. What you get as the y value after you do out the equation is r.

    Hope this helps!
    I'm sorry I don't understand.


  • Registered Users, Registered Users 2 Posts: 6,840 ✭✭✭Luno


    Moromaster wrote: »
    The teacher won't explain it again as I was absent, how do you do this question:

    L is the line 3x-y-11=0. The line K a(-3, 9) and b(6, r). K is perpendicular to L. Find R.
    Moromaster wrote: »
    I'm sorry I don't understand.

    To find r you need to sub in values for x and y in the original equation so...
    (6,r) =>
    3(6)-(r)-11=0
    18-r-11=0
    7-r=0
    -r=-7
    r=7

    Could be wrong...

    Actually I don't get how the whole perpendicular thing is relevant to this question so I would say I'm wrong ... Is there more to the question?

    Can anyone do this cause I'm anxious now!


  • Closed Accounts Posts: 1,207 ✭✭✭Ally7


    Moromaster wrote: »
    The teacher won't explain it again as I was absent, how do you do this question:

    L is the line 3x-y-11=0. The line K a(-3, 9) and b(6, r). K is perpendicular to L. Find R.

    That's ok, just realised now that i explained wrong!

    First, you must know that when you multiply the slopes of two perpendicular lines, the answer will equal -1. To work this out if you get the slope of one the lines, you invert it and change the sign to get the slope of your other perpendicular line.
    Once you've found the slope of L, invert it and change the sign and this is the slope of K.
    Next, you need to find the equation of the line K. you can do this with the formula y-y1=m(x-x1), with m being your slope of L and x1=-3 and y1=9 , coming from the point a above.
    Once you've got the equation of the line L, you can now find r. If you sub in 6 for x in the equation and work it out, you will eventually get y=?, and ? will be the value of r.


  • Closed Accounts Posts: 777 ✭✭✭Mayoegian


    Moromaster wrote: »
    The teacher won't explain it again as I was absent, how do you do this question:

    L is the line 3x-y-11=0. The line K a(-3, 9) and b(6, r). K is perpendicular to L. Find R.

    Hi there!:)

    Because K is perpendicular to L, find the slope of L, turn it upside down and change the sign to obtain the slope of K. You then find the slope of K, in terms of 'r' from the two points you are given. Equal this result to the slope of K, and you should solve to find 'r.'


    This is how you do it:

    1) Line L: 3X -Y-11=0
    Therefore, slope of L is,
    Y=3X - 11
    Which is 3. So, the slope of K is -1/3
    When finding the slope of a given line, make Y equal the X and other value, then take the number in front of the X as your slope.

    2) Points on K: (-3,9) (6,r)
    Find slope using the formula:
    Y2-Y1/X2-X1
    You should get r-9/9

    3) You then write your equation like this:

    r-9/9 = -1/3
    Solving for r you should get r=6!


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