Find the distance ##d## to a point in the given problem

[chwala]

Homework Statement

see attached.

Relevant Equations

distance formula

My problem is on how to arrive at $d=\dfrac{mx-y}{\sqrt{1+m^2}}$

My working steps are as follows;

$d^2=(x_1 - x)^2+ (y_1-y)^2$

$d^2=(\dfrac{y}{m} -x)^2+ (mx-y)^2$

$d^2=\dfrac{(mx-y)^2}{m^2} + (mx-y)^2$

$m^2d^2=(mx-y)^2(1+m^2)$

$d=\dfrac{mx-y\sqrt{1+m^2}}{m}$

...unless they made a mistake!

 

[Mark44]

There's a formula and proof for the perpendicular distance of a point to a line here: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line.

I'm having a hard time following your work right from the start. The drawing shows a point (x, y) and another point (x', y'). In your work you have $x_1$ and y. Shouldn't you have $y_1$?

 

[chwala]

There's a formula and proof for the perpendicular distance of a point to a line here: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line.

I'm having a hard time following your work right from the start. The drawing shows a point (x, y) and another point (x', y'). In your work you have $x_1$ and y. Shouldn't you have $y_1$?

If there is a formula then that is fine. I will need to look at it.

I used the distance formula,

$d=\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$ with my points being, $(\frac{y}{m}, mx)$ and $(x,y)$ the rest of the working (check my second line) ought to be clear to you sir.

 

[chwala]

I can see my mistake, i failed to take into consideration the intersection points between the line and the external point.

We have to determine the points of intersection between the external point and our line $y=mx$ we shall use the normal equation of the line $\left[\dfrac{y-y_0}{x-x_0}\right] =\left[ \dfrac{-1}{m}\right]$... that is perpendicular to the straight line $y=mx$ this will eventually lead us to the final equation, $d=\dfrac{|k+mx_0-y_o|}{\sqrt{1+m^2}}$ , our line equation being $y=mx+0$, the result would be $d=\dfrac{mx-y}{\sqrt{1+m^2}}$ which is as shown on the text.

 

[chwala]

So, is it an assumption that the external point shall be perpendicular to the given straight line? supposing the point is not perpendicular to the given line?

 

[SammyS]

So, is it an assumption that the external point shall be perpendicular to the given straight line? supposing the point is not perpendicular to the given line?

What do you mean, "the external point"? Do you mean the point, P ? It has a label so that we can refer to it by name.

Furthermore, how is it that a point is or is not perpendicular to a line?

To find the distance of a point from a line, you measure the distance along a line which is perpendicular to the given line. Right?

All pretty basic stuff.

 

[Lnewqban]

So, is it an assumption that the external point shall be perpendicular to the given straight line? supposing the point is not perpendicular to the given line?

Consider d to be the radius of a circle to which the line is a tangent.
The tangent to the circle is perpendicular to the radius of the circle at the point of contact.

In this case, we have two circles of equal radii d, intersecting at a single point with a line that is tangent to both.
The distance between the centers of those circles is 2d.

 

[WWGD]

There are results to the effect that the Projection onto the line will provide the shortest distance.

 

[chwala]

There are results to the effect that the Projection onto the line will provide the shortest distance.

Thanks @WWGD ,its something that i know, i guess i overthink too much at times. Cheers man.