Thread: RE: Could you proof it?

1. RE: Could you proof it?

Hai, everyone..once again, its me, Ah Chan. Sorry for I'm quite inactive this last few weeks because I got lot of problems banging my doors and so busy, rushing the assignment...anyways, thanks to everyone for my previous last questions. Erm, I got a proofing question here that had took me quite a long times yet I haven't fully understand the solution for it, mind to help me out...thanks a lot anyways

Prove for distance from a point to line formula;

First, show that the distance from O (0,0) to Ax+By+C=0 is lCl / (square root)[A^2 + B^2] by finding the area of the shaded triangle using two different altitudes and then equating the areas....(Hints: Draw a right-angle triangle with two points, [0,(-C/B)] and [(-C/A),0]

Then, now translate (h,k) and the line Ax+By+C=0 by the same amount so that (h,k) moves to (0,0)...

The answer should be d = lAh+Bk+Cl / (square root)[A^2+B^2]..but I just couldn't figured it out, please do help me..thanks

2. Originally Posted by Ah Chan
Hai, everyone..once again, its me, Ah Chan. Sorry for I'm quite inactive this last few weeks because I got lot of problems banging my doors and so busy, rushing the assignment...anyways, thanks to everyone for my previous last questions. Erm, I got a proofing question here that had took me quite a long times yet I haven't fully understand the solution for it, mind to help me out...thanks a lot anyways

Prove for distance from a point to line formula;

First, show that the distance from O (0,0) to Ax+By+C=0 is lCl / (square root)[A^2 + B^2] by finding the area of the shaded triangle using two different altitudes and then equating the areas....(Hints: Draw a right-angle triangle with two points, [0,(-C/B)] and [(-C/A),0]

Then, now translate (h,k) and the line Ax+By+C=0 by the same amount so that (h,k) moves to (0,0)...

The answer should be d = lAh+Bk+Cl / (square root)[A^2+B^2]..but I just couldn't figured it out, please do help me..thanks
First I am not sure what area has to do with distance in the problem....

Key point 1. When we talk about the distance between a point and a line it means the perpendicular distance.

With that in mind here we go.

$Ax+By+C=0 \iff y=-\frac{A}{B}x-\frac{C}{B}$

So the slope of this line is $m=-\frac{A}{B}$

So the slope of the perpendicular line is $m_\perp=\frac{B}{A}$

So the equation of the line is $y=\frac{B}{A}x$

Now we want to find the intersection of these two lines so we set them equal.

$\frac{B}{A}x=-\frac{A}{B}x-\frac{C}{B}$ clearing the fractions we get

$B^2x=-A^2x-AC \iff x=\frac{-AC}{A^2+B^2}$

Now we can sub this into either equation of the line to get the y coordinate

$y=\frac{B}{A}\cdot \frac{-AC}{A^2+B^2}=\frac{-BC}{A^2+B^2}$

Now we can use the pythagorean theorem to find the distance between the two lines.

$\left( \frac{-AC}{A^2+B^2} \right)^2+\left( \frac{-BC}{A^2+B^2} \right)^2=d^2$

Simplifing we get

$\frac{C^2(A^2+B^2)}{(A^2+B^2)^2}=d^2$

$\frac{|C|}{\sqrt{A^2+B^2}}=d$

This should get you started. Good luck.