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

- Sep 13th 2008, 05:36 PMgobbajeezalusWhat 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

- Sep 13th 2008, 05:56 PMChop Suey
Let's call line $\displaystyle y = \frac{1}{2}x + 6$: 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:

$\displaystyle y - y_1 = m(x - x_1)$

Find the point of intersection of these 2 lines. The perpendicular distance from the pt 5, 4 to the line $\displaystyle y = \frac{1}{2}x + 6$ is the distance from (5,4) to the point of intersection. - Sep 13th 2008, 06:10 PMSoroban
Hello, gobbajeezalus!

Quote:

What is the perpendicular distance from $\displaystyle P(5,4)$ to the line $\displaystyle L_1\!:\;y \:= \:\frac{1}{2}x + 6$ ?

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

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

$\displaystyle L_2$ has point (5,4) and slope -2

. . Its equation is: .$\displaystyle y - 4 \:=\:-2(x - 5) \quad\Rightarrow\quad y \:=\:-2x + 14$

Where do $\displaystyle L_1$ and $\displaystyle L_2$ intersect?

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

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

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

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

- Sep 13th 2008, 06:28 PMShyam
$\displaystyle {\text{Perpendicular distance from a point }}$

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

$\displaystyle = \frac{{\left| {ax_1 + by_1 + c} \right|}}

{{\sqrt {a^2 + b^2 } }} \hfill \\$

$\displaystyle {\text{So, perpendicular distance from }}\left( {5,4} \right){\text{ to line }}\frac{1}

{2}x - y + 6 = 0{\text{ is given as:}} \hfill \\$

$\displaystyle = \frac{{\left| {\frac{1}

{2}\left( 5 \right) + \left( { - 1} \right)\left( 4 \right) + 6} \right|}}

{{\sqrt {\left( {\frac{1}

{2}} \right)^2 + \left( { - 1} \right)^2 } }} = \frac{{4.5}}

{{\sqrt {1.25} }} \approx 4.025 \hfill \\

$