Results 1 to 2 of 2

Math Help - lines perpendicular

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517

    lines perpendicular

    hey guys, need help with this .. any suggestions with how i should start with question? im not looking for the answer, i want to learn how to do it..thank u

    The lines given by
    ax + by = 3 and cx 4y = 5 are perpendicular and intersect at (6, 7). Find a, b, and c.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by jvignacio View Post
    hey guys, need help with this .. any suggestions with how i should start with question? im not looking for the answer, i want to learn how to do it..thank u

    The lines given by
    ax + by = 3 and cx 4y = 5 are perpendicular and intersect at (6, 7). Find a, b, and c.
    We have the lines ax+by=3 and cx-4y=5

    They are perpendicular and intersect at (6,7)

    Recall that the slopes of perpendicular lines are opposite reciprocals of each other.

    So we want to find \frac{\,dy}{\,dx} of both lines.

    So we see that \frac{\,dy}{\,dx}=-\frac{a}{b} and \frac{\,dy}{\,dx}=\frac{c}{4}

    Now, let us evaluate the lines at the intersection point:

    6a+7b=3-----(1)
    6c=33--------(2)

    We can now solve for c:

    6c=33\implies c=\dots

    Now substitute this value into the derivative.

    \frac{\,dy}{\,dx}=\frac{c}{4}

    We see that the negative reciprocal of this slope will give us a slope parallel to the other line. We want these slopes to be equal.

    Thus, we must let -\frac{a}{b}=-\frac{4}{c}

    We must solve this system of equations in order to find a and b.

    \left\{\begin{array}{r}6a+7b=3\\ \frac{a}{b}=\frac{4}{c}\end{array}\right.

    I hope this makes sense!

    --Chris
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with Perpendicular Lines
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 13th 2009, 01:47 PM
  2. Perpendicular lines
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 1st 2008, 11:55 AM
  3. Perpendicular Lines
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 11th 2008, 07:53 PM
  4. Perpendicular lines
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 21st 2007, 04:28 AM
  5. Perpendicular Lines
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 26th 2007, 03:22 PM

Search Tags


/mathhelpforum @mathhelpforum