Sologuitar's question at Yahoo Answers (perpencicular distance)

[Fernando Revilla]

Here is the question:

Show that the perpendicular distance from the origin to the line y = mx + b is the absolute value of b divided by the square root of the quantity (1 + m^2).

Here is a link to the question:

Perpendicular Distance From (0,0)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello sologuitar,

In general the distance from the point $P_0=(x_0.y_0)$ to the line $r:Ax+By+C=0$ is

$$d(P_0,r)=\left|\dfrac{Ax_0+By_0+C}{\sqrt{A^2+B^2}}\right|$$

In our case, $P_0=(0,0)$ and $r:mx-y+b=0$ so,

$$d(P_0,r)=\left|\dfrac{m\cdot 0+(-1)\cdot 0+b}{\sqrt{m^2+(-1)^2}}\right|=\dfrac{\left|b\right|}{\sqrt{1+m^2}}$$

 

[MarkFL]

Hello sologuitar,

For a derivation of a similar formula to that cited by Fernando Revilla, see this topic:

http://www.mathhelpboards.com/f49/finding-distance-between-point-line-2952/
Mark.

 

[soroban]

Hello, all!

I'll solve this "from scratch".

$\text{Show that the perpendicular distance from the origin to the line }L:\;y \,=\,mx+b$
$\text{is the absolute value of }b\text{ divided by the square root of }(1+m^2).$

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The line $L$ has slope $m$.
The line through the origin has slope $\text{-}\frac{1}{m}$.

Point $P$ is the intersection of lines $y \:=\:mx+b$ and $y \:=\:\text{-}\frac{1}{m}x$

. . $mx + b \:=\:\text{-}\dfrac{1}{m}x \quad\Rightarrow\quad mx + \dfrac{1}{m}x \:=\:\text{-}b$

. . $\dfrac{m^2+1}{m}x \:=\:\text{-}b \quad\Rightarrow\quad x \:=\:\dfrac{\text{-}bm}{1+m^2}$

Then $y \:=\:\text{-}\dfrac{1}{m}\left(\dfrac{\text{-}bm}{1+m^2}\right) \quad\Rightarrow\quad y \:=\:\dfrac{b}{1+m^2} $

Point $P$ has coordinates: $\left(\dfrac{\text{-}bm}{1+m^2},\:\dfrac{b}{1+m^2}\right)$Distance $OP$ is: .$d \;=\;\sqrt{\left(\dfrac{\text{-}bm}{1+m^2}\right)^2 + \left(\dfrac{b}{1+m^2}\right)^2} \;=\; \sqrt{\dfrac{b^2m^2}{(1+m^2)^2} + \dfrac{b^2}{(1+m^2)^2}} $

. . . . . . . . . . . $d \;=\;\sqrt{\dfrac{b^2(1+m^2)}{(1+m^2)^2}} \;=\;\sqrt{\dfrac{b^2}{1+m^2}} $

. . Therefore: .$d \;=\;\dfrac{|b|}{\sqrt{1+m^2}} $

 

[CellCoree]

Hello Sologuitar,

Thank you for your question. The proof for the formula you mentioned can be shown using basic geometry and the Pythagorean Theorem.

Let's start by drawing a coordinate system with the origin (0,0) and the line y = mx + b. We know that the distance from the origin to any point on the line is the shortest distance, and this distance is perpendicular to the line.

Now, let's choose a point on the line with coordinates (x, mx + b). The distance from this point to the origin can be calculated using the Pythagorean Theorem:

d^2 = (x - 0)^2 + (mx + b - 0)^2
d^2 = x^2 + (mx + b)^2

Next, we want to find the minimum distance, which occurs when the point (x, mx + b) is perpendicular to the line. This means that the slope of the line connecting the origin to this point is the negative reciprocal of the slope of the given line, which is -1/m. Using this information, we can write the equation for this line as:

y = -1/mx + c

where c is the y-intercept of this line. Since this line passes through the point (x, mx + b), we can solve for c:

mx + b = -1/mx + c
c = mx + b + 1/mx

Now, let's substitute this value of c into our previous equation for d^2:

d^2 = x^2 + (mx + b)^2
d^2 = x^2 + (mx + b + 1/mx)^2

Next, we want to find the minimum value of d^2, which occurs when the derivative of d^2 with respect to x is equal to 0. Taking the derivative, we get:

d^2 = 2x + 2(mx + b + 1/mx)(m + 1/m)
d^2 = 2x + 2(m^2x + b + 1/m)
d^2 = 2x + 2m^2x + 2b + 2/m

Setting this equal to 0 and solving for x, we get:

x = -b/(1+m^2)

Substituting this value of x back into our equation for d^