# Thread:

1. ## 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

2. 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.

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

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

4. Originally Posted by gobbajeezalus
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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