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Math Help - What is the perpendicular distance from the pt 5, 4 to the line y = 1/2x + 6?

  1. #1
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    What is the perpendicular distance from the pt 5, 4 to the line y = 1/2x + 6?

    The title pretty much explains the question:

    What is the perpendicular distance from the pt 5, 4 to the line y = 1/2x + 6?

    If you could please tell the answer and explain how you got there that would be great. thanks
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  2. #2
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    Let's call line y = \frac{1}{2}x + 6: line k.

    Find the equation of the line whose slope is the negative reciprocal of line K and it includes the point (5, 4). Call this line J. You will need to use the point-slope form equation:
    y - y_1 = m(x - x_1)

    Find the point of intersection of these 2 lines. The perpendicular distance from the pt 5, 4 to the line y = \frac{1}{2}x + 6 is the distance from (5,4) to the point of intersection.
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  3. #3
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    Hello, gobbajeezalus!

    What is the perpendicular distance from P(5,4) to the line L_1\!:\;y \:= \:\frac{1}{2}x + 6 ?
    There is a formula for this problem, but I'll walk through it for you . . .

    The given line (L_1) has slope \frac{1}{2}

    The line perpendicular to it (L_2) has slope -2

    L_2 has point (5,4) and slope -2
    . . Its equation is: . y - 4 \:=\:-2(x - 5) \quad\Rightarrow\quad y \:=\:-2x + 14


    Where do L_1 and L_2 intersect?

    . . \frac{1}{2}x + 6 \;=\;-2x + 14 \quad\Rightarrow\quad \frac{5}{2}x \:=\:8

    Hence: . x \:=\:\frac{16}{5},\;\;y \:=\:\frac{38}{5}\quad\hdots . They intersect at: . Q\left(\frac{16}{5},\:\frac{38}{5}\right)


    The desired distance is: . PQ \;=\;\sqrt{\left(\frac{16}{5} - 5\right)^2 + \left(\frac{38}{5} - 4\right)^2} \;= \;\sqrt{\left(\text{-}\frac{9}{5}\right)^2 + \left(\frac{18}{5}\right)^2}


    . . = \;\sqrt{\frac{81}{25} + \frac{324}{25}} \;=\;\sqrt{\frac{405}{25}} \;=\;\sqrt{\frac{81\cdot5}{25}} \;=\;\boxed{\frac{9\sqrt{5}}{5} \;\approx\;4.025}

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  4. #4
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    Quote Originally Posted by gobbajeezalus View Post
    The title pretty much explains the question:

    What is the perpendicular distance from the pt 5, 4 to the line y = 1/2x + 6?

    If you could please tell the answer and explain how you got there that would be great. thanks
    {\text{Perpendicular distance from a point }}

    \left( {x_1 ,y_1 } \right){\text{ to the line }}ax + by + c = 0{\text{ is given by the formula:}} \hfill \\

      = \frac{{\left| {ax_1  + by_1  + c} \right|}}<br />
{{\sqrt {a^2  + b^2 } }} \hfill \\

    {\text{So, perpendicular distance from }}\left( {5,4} \right){\text{ to line }}\frac{1}<br />
{2}x - y + 6 = 0{\text{ is given as:}} \hfill \\

       = \frac{{\left| {\frac{1}<br />
{2}\left( 5 \right) + \left( { - 1} \right)\left( 4 \right) + 6} \right|}}<br />
{{\sqrt {\left( {\frac{1}<br />
{2}} \right)^2  + \left( { - 1} \right)^2 } }} = \frac{{4.5}}<br />
{{\sqrt {1.25} }}  \approx  4.025 \hfill \\ <br />
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