Results 1 to 3 of 3

Math Help - Deceptively simple looking . . .

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    4

    Question Deceptively simple looking . . .

    Hello! I'm kind of new to this forum, and I hope I'm posting in the right category. My school calls the class I'm taking "College Algebra", but I've been told by some that it's roughly equal to Pre-Calc.

    My question on this worksheet looks simple on the surface, but I never seem to get the right answer (-4/5) and it'd be nice to know why. Here it is:

    Find the value of a if the line through (-1, a) and (3, -4) is parallel to y=ax.

    What I've been doing is plugging (3, -4) into the equation as the x- and y-values, but that gets me the answer -4/3, which is not quite right.

    Any suggestions?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by potato92 View Post
    Hello! I'm kind of new to this forum, and I hope I'm posting in the right category. My school calls the class I'm taking "College Algebra", but I've been told by some that it's roughly equal to Pre-Calc.

    My question on this worksheet looks simple on the surface, but I never seem to get the right answer (-4/5) and it'd be nice to know why. Here it is:

    Find the value of a if the line through (-1, a) and (3, -4) is parallel to y=ax.

    What I've been doing is plugging (3, -4) into the equation as the x- and y-values, but that gets me the answer -4/3, which is not quite right.

    Any suggestions?
    First note the point is not on the line. However the slope of the line is m=a

    Also since you have two points you can use the relation

    m=\frac{y_2-y_1}{x_2-x_1}

    Now just set the two eqaul and solve for a.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466
    Yes, "plugging" the points into the equation y= ax will not give you anything because those points are NOT on that line- they are on a line parallel to that line.

    Another way to do this- any non-vertical line can be written in the form y= mx+ b where "m" is the slope and b is the y-intercept. The slope of line y= ax is a and any line parallel to it will have slope a also. So you know your line has equation y= ax+ b.

    Put x= 3, y= -4 into that equation to determine and equation for a and b. Put x= -1, y= a to get a second equation for a and b.

    solve those equations for a and b and write out the equation of the line.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Really simple question too hard for a simple mind
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 5th 2010, 08:03 AM
  2. Easy Question - Simple Interest, Simple Discount
    Posted in the Business Math Forum
    Replies: 0
    Last Post: September 21st 2010, 08:22 PM
  3. Replies: 4
    Last Post: May 4th 2010, 09:08 AM
  4. Replies: 2
    Last Post: November 12th 2009, 03:09 PM
  5. Deceptively difficult trig problem
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 28th 2008, 08:48 PM

Search Tags


/mathhelpforum @mathhelpforum