Results 1 to 3 of 3

Math Help - equidistant points

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    equidistant points

    Our topic is about distance formulas:

    1.) If the point (x,3) is equidistant from (3,-2) and (7,4) find x

    2.) Find the point on the y - axis which is equidistant from (-5,-2) and (3,2)

    thank you very much
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    nr. 1.) only

    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Our topic is about distance formulas:

    1.) If the point (x,3) is equidistant from (3,-2) and (7,4) find x...
    Hello, Engineer Adam,

    the point you are looking for belongs to the perpendicular bisector of the two given points and the straight line parallel to the x-axis with y = 3.

    1. Calculate the midpoint between the given points: M(5, 1)
    2. Calculate the slope of the line connecting the given points:
    m=\frac{4-(-2)}{7-3}=\frac{3}{2}

    Therefore the perpendicular direction has the slope -2/3

    3. Use point-slope-formula of a straight line:

    \frac{y-1}{x-5}=-\frac{2}{3} \ \text{and} \ y=3

    4. Plug in this value for y and solve for x. You should get x = 2

    So the point you are looking for is P(2, 3)

    EB

    PS.: The 2nd problem is very similar to this one here. I leave for you.
    Attached Thumbnails Attached Thumbnails equidistant points-gl_entferng1.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644
    Hello, ^_^Engineer_Adam^_^!

    Our topic is about distance formulas. . Really?

    Here's an idea . . . How about using the Distance Formula?

    . . . . . d \;=\;\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}
    1) If the point P(x,3) is equidistant from A(3,-2) and B(7,4) find x

    Distance \overline{PA} \:=\:\sqrt{(x-3)^2 + (3-[-2])^2} \:=\:\sqrt{x^2-6x + 34}

    Distance \overline{PB}\:=\:\sqrt{(x-7)^2 + (3-4)^2}\:=\:\sqrt{x^2-14x + 50}

    Since \overline{PA} = \overline{PB}, we have: . \sqrt{x^2-6x+34} \:=\:\sqrt{x^2-14x+50}

    Square both sides: . x^2 - 6x + 34 \:=\:x^2-14x + 50

    And we have: . 8x \:=\:16\quad\Rightarrow\quad \boxed{x\,=\,2}



    2) Find the point on the y - axis which is equidistant from A(-5,-2) and B(3,2)

    A point on the y-axis has the form: P(0,\,y)

    Then: . \begin{array}{cc}\overline{PA} \:= \:\sqrt{(0+5)^2+(y+2)^2} \:= \:\sqrt{y^2+4y+29} \\ \overline{PB} \:= \:\sqrt{(0-3)^2 + (y-2)^2} \:= \:\sqrt{y^2 - 4y + 13} \end{array}

    Hence: . \sqrt{y^2 + 4y + 29} \:=\:\sqrt{y^2 - 4y + 13}\quad\Rightarrow\quad y^2 + 4y + 29 \:=\:y^2 - 4y + 13

    Therefore: . 8y\:=\:-16\quad\Rightarrow\quad y \:=\:-2 . . The point is \boxed{(0,\,-2)}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Points Equidistant
    Posted in the Algebra Forum
    Replies: 6
    Last Post: November 3rd 2010, 03:44 PM
  2. Three Points Equidistant...
    Posted in the Geometry Forum
    Replies: 2
    Last Post: September 19th 2010, 07:27 PM
  3. Find plane equidistant between two points
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 14th 2010, 11:01 AM
  4. Point equidistant to 3 or more other points
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: October 30th 2009, 02:27 AM
  5. Coordinates of linked equidistant points on R3
    Posted in the Geometry Forum
    Replies: 0
    Last Post: April 29th 2009, 06:00 PM

Search Tags


/mathhelpforum @mathhelpforum