A Dot product analysis proof (might be simple)

[Andrax]

Homework Statement

so we have a (D) Line (geometry) it's Cartesian equation is ax+by+c=0
we have an [itex]A[/itex]([itex]\alpha[/itex],[itex]\beta[/itex])
prove that the distance between the line(D) and the point A is
d=[itex]\frac{la\alpha+b\beta+cl}{\sqrt{a^2+b^2}}[/itex]

Homework Equations

The Attempt at a Solution

let every distance be a vector (sorry didn't find vectors in the latex reference)
AH h belongs to (D) and AH factors (D) so AH(a,b)
we also have n vector (-b,a)
we can say that the catersian equation of ah is -b(x-[itex]\alpha[/itex])+a(y-[itex]\beta[/itex])+c I am pretty musch stuck here now to get this \sqrt{a^2+b^2} we will probably use cos but ireally can't find how r we supposed to use it :( please help guys:

 

[mfb]

What are H and h? What is l?
Do you know the cross-product?
Alternatively, can you transform your line to get ##\sqrt{a^2+b^2}=1##? This is always possible (as a!=0 or b!=0 for a meaningful line).

 

[HallsofIvy]

The distance from a point to a line is, by definition, measured along a line through the given point perpendicular to the given line.

We can write ax+ by= c as y= -(a/b)x+ c/b showing that the line has slope -(a/b). A line perpendicular to it must have slope b/a. A line with that slope and passing through the point [itex](\alpha, \beta)[/itex] has equation [itex]y= (b/a)(x- \alpha)+ \beta[/itex] or [itex]bx- ay= b\alpha- a\beta[/itex]. Determine where the two lines intersect, by solving ax+ by= c and [itex]bx- ay= b\alpha- a\beta[/itex] simultaneously and then determine the distance between that point of intersection and [itex](\alpha, \beta)[/itex].

That should get the formula you want.

 

[Andrax]

The distance from a point to a line is, by definition, measured along a line through the given point perpendicular to the given line.

We can write ax+ by= c as y= -(a/b)x+ c/b showing that the line has slope -(a/b). A line perpendicular to it must have slope b/a. A line with that slope and passing through the point [itex](\alpha, \beta)[/itex] has equation [itex]y= (b/a)(x- \alpha)+ \beta[/itex] or [itex]bx- ay= b\alpha- a\beta[/itex]. Determine where the two lines intersect, by solving ax+ by= c and [itex]bx- ay= b\alpha- a\beta[/itex] simultaneously and then determine the distance between that point of intersection and [itex](\alpha, \beta)[/itex].

That should get the formula you want.

thank you.

 

[Andrax]

What are H and h? What is l?
Do you know the cross-product?
Alternatively, can you transform your line to get ##\sqrt{a^2+b^2}=1##? This is always possible (as a!=0 or b!=0 for a meaningful line).

thank you but i we didn't study that "theory" so i don't think I'm allowed to apply it anyway the hallsofivy provided a rather simple solution .

 

[Ray Vickson]

thank you but i we didn't study that "theory" so i don't think I'm allowed to apply it anyway the hallsofivy provided a rather simple solution .

Strictly speaking, the distance between a line and a point is defined to be the *shortest* distance between the given point and points on the line. It follows (basically, as a theorem) that the shortest line is perpendicular to the given line, as HallsOfIvy has stated.

 

[HallsofIvy]

Strictly speaking, the distance between a line and a point is defined to be the *shortest* distance between the given point and points on the line. It follows (basically, as a theorem) that the shortest line is perpendicular to the given line, as HallsOfIvy has stated.

In fact, it follows from the Pythagorean theorem. If we measure from point P to line l along a NON perpendicular line, we can drop a perpendicular from P to l giving us a right triangle in which the original, non-perpendicular, distance is the hypotenuse. By [itex]c^2= a^2+ b^2[/itex], c, the length of the hypotenuse is longer than the length of either leg.

Conversely, you can define the "distance between point P and line l" to be measured along the perpendicular and use that theorem to show that it is the shortest distance.