Show that the distance between the points (x,y) and the line Ax + By + C = 0 is:
D = (Ax + By + C) / sqrt(A^2 + B^2)
So, would you start out by multiplying sqrt (A^2 + B^2) to both sides and then it would become:
sqrt (A^2 + B^2)D = Ax + By + C
Then square both sides,
D^2(A^2 + B^2) = (Ax + By + C)^2
Then I got lost, could someone direct me through this problem? Did I start out the right way? Thanks!
Just manipulating the equation for D will not constitute a proof that it is correct. You need to derive it somehow.
Try deriving an equation for a line through (x,y) that is perpendicular to the line Ax + By +C = 0. Then compute where the two lines intersect. Call that point (x', y'). Then compute the distance between (x,y) and (x',y').
A proof can be found here:
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
To derive the formula for the distance between a point (x,y) and a line Ax + By + C = 0, you need to use the concept of perpendicular distance. Here's how you can proceed:
1. Start with the equation of the line: Ax + By + C = 0.
2. Consider a point (x, y) that lies off the line.
3. The perpendicular distance (D) between this point and the line is the shortest distance.
4. Consider a line passing through (x, y) and perpendicular to the given line.
5. Find the equation of this perpendicular line.
5.1 The slope of the given line is -A/B (since it's perpendicular, the slopes will be negative reciprocals).
5.2 The slope of the perpendicular line will be B/A.
5.3 Using the point-slope form, the equation of the perpendicular line passing through (x, y) is: y - y = (B/A)(x - x), which simplifies to Bx - Ay + Ay - Bx = 0. Rearranging gives Bx - Ay = C' (let's call it C') as the equation of the perpendicular line.
6. Determine the point of intersection between the original line Ax + By + C = 0 and the perpendicular line Bx - Ay = C'.
6.1 Solve these equations simultaneously to find the coordinates of the point of intersection (x', y').
7. The distance (D) between (x, y) and (x', y') is the shortest distance, which is the perpendicular distance.
8. Use the distance formula:
D = √[(x - x')^2 + (y - y')^2].
Now let's substitute the equation of the perpendicular line (Bx - Ay + C') into the distance formula:
D = √[(x - x')^2 + (y - y')^2]
= √[(x - x')^2 + (y - y')^2]
= √[(x - x')^2 + (y - (Ax' + By' + C')/A)^2],
where we substitute the value of y from the equation of the perpendicular line into y - y':
= √[(x - x')^2 + (y - Ax' - By' - C'/A)^2],
Simplifying further:
D = √[(x - x')^2 + (y - Ax' - By' - C'/A)^2].
Comparing this equation with the given equation of the line Ax + By + C = 0, we can see that:
- The coefficient of x' is A.
- The coefficient of y' is B.
- The constant term is C'/A, which can be written as C/A since C' = C.
Therefore, the equation for the distance D becomes:
D = √[(x - x')^2 + (y - Ax - By - C/A)^2],
or
D = √[(x - x')^2 + (y - Ax - By - C)^2/(A^2 + B^2)].
Finally, we simplify the last equation by multiplying both sides by √(A^2 + B^2):
D√(A^2 + B^2) = √[(x - x')^2 + (y - Ax - By - C)^2].
Squaring both sides:
D^2(A^2 + B^2) = (x - x')^2 + (y - Ax - By - C)^2.
This equation relates the distance (D) between the point (x, y) and the line Ax + By + C = 0 to the coordinates of the point of intersection (x', y'). You can rewrite it as:
D = √[(x - x')^2 + (y - Ax - By - C)^2]/√(A^2 + B^2),
which can be further simplified as:
D = |Ax + By + C|/√(A^2 + B^2).
Hence, the formula for the distance (D) between a point (x, y) and the line Ax + By + C = 0 is:
D = |Ax + By + C|/√(A^2 + B^2).
To summarize, you started off correctly, but there were a few steps missing. Follow the above steps, and you will derive the required formula for the distance between a point and a line.