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Thread: Re: find the distance of a point from a line?

  1. #1
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    Re: find the distance of a point from a line?

    use the following procedure to find the leats ( perpendicular ) distance of the point ( 1,2 ) from the line y = 3x + 5, without having to find the equation of a line perpendicular to y = 3x + 5.

    (a) let ( x,y ) be a general point on the line, Show that its distance, d, from ( 1,2 ) is given by d^2 = ( x - 1 )^2 + ( y - 2 )^2.

    (b) use the equation of the line to show that d^2 = ( x - 1 )^2 + ( 3x 3 )^2. I CAN SEE THE ANSWER HERE SO NO WORRIES.

    (c) by completing the square, show that the minimum distance required 3/5√10
    Can you please solve this question completely ?
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  2. #2
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    Re: find the distance of a point from a line?

    Quote Originally Posted by Isra View Post
    use the following procedure to find the leats ( perpendicular ) distance of the point ( 1,2 ) from the line y = 3x + 5, without having to find the equation of a line perpendicular to y = 3x + 5.

    (a) let ( x,y ) be a general point on the line, Show that its distance, d, from ( 1,2 ) is given by d^2 = ( x - 1 )^2 + ( y - 2 )^2.

    (b) use the equation of the line to show that d^2 = ( x - 1 )^2 + ( 3x 3 )^2. I CAN SEE THE ANSWER HERE SO NO WORRIES.

    (c) by completing the square, show that the minimum distance required 3/5√10
    Can you please solve this question completely ?
    surely you can do (a), it's just applying the definition of the distance between two points in Euclidean space.

    you say you can do (b), ok good. but for good form

    $d^2 = (x-1)^2 + (y-2)^2$

    $d^2 = (x-1)^2 + (3x+5-2)^2$

    $d^2 = (x-1)^2 + (3x+3)^2$ (ah, you're missing a "+" sign)

    $d^2 = x^2 - 2x + 1 +9x^2 + 18x+9$

    $d^2 = 10x^2 + 16x + 10$

    can you go ahead and complete the square and find the mininum distance now?
    Thanks from skeeter
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