Lines in a plane - Orthogonality; Distances
orthonormal-axes
In all that follows we assume an X-axis orthogonal to a Y-axis fixed in o and so that the unities on both axes have the same magnitude.
We call this : 'orthonormal axes'.
We know that, in a plane, each line l has an equation of the form
ax + by + c = 0
If b is not 0, the slope of that line is -a/b .
Remark: ax + by = 0 is the equation of a line parallel to l and containing the origin o.
Let line l : ax + by + c = 0 and l' : ax + by = 0
Each point P on l' is the image point of a vector P defining the direction of l. Then, P is called a direction vector of l.
If r is a real number (not 0 ) , then r.P is a direction vector too.
A simple choice for P is P(b,-a).
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(b,-a) are the coordinates of a direction vector of the line ax + by + c = 0
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Take two lines l : ax + by + c = 0 and l' : a'x + b'y + c' = 0
Direction vectors are P(b,-a) and Q(b',-a').
Now,
l and l' are orthogonal
<=>
P and Q are orthogonal
<=>
P.Q = 0
<=>
b.b' + a.a' = 0
<=>
a.a' + b.b' = 0
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Two lines l : ax + by + c = 0 and l' : a'x + b'y + c' = 0 are orthogonal if and only if
a.a' + b.b' = 0 .
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Take two lines l : ax + by + c = 0 and l' : a'x + b'y + c' = 0
If b and b' are not zero, the slope
is m = -a/b and m' = -a'/b' .
Now,
l and l'are orthogonal
<=>
a.a' + b.b' = 0
<=>
aa'
---- + 1 = 0
bb'
<=>
m.m' + 1 = 0
<=>
m.m' = -1
Line l with slope m and line l' with slope m'
are orthogonal if and only if m.m' = -1.
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Let l : ax + by + c = 0
The vector P(b,-a) is a direction vector of l.
The vector Q(a,b) is orthogonal with P because P.Q = 0.
Hence, Q(a,b) is a vector orthogonal to l.
We call Q(a,b) a normal vector to l.
Let l : ax + by + c = 0
Then also l : rax + rby + rc = 0 with r a real number not 0.
We calculate a suitable r such that the normal vector Q(ra,rb)
is a unity vector.
Q(ra,rb) is a unity vector
<=>
Q.Q = 1
<=>
ra.ra + rb.rb = 1
<=>
r2 (a2 + b2 ) = 1
<=>
1 - 1
r = ------------ or r = ------------
___________ ____________
V a2 + b2 V a2 + b2
Substitute the positive value of r in the equation of l.
ax + by + c
l : ----------------- = 0
____________
V a2 + b2
This equation is called a normal equation of l.
Hence,
lx + my + n = 0 is a normal equation of line l.
<=>
Q(l,m) is a unity normal vector to l
<=>
l2 + m2 = 1
l : lx + my + n = 0 is a normal equation of l.
<=>
l2 + m2 = 1
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Take a point P(a,b) and a line l with normal equation lx + my + n = 0 .
Then Q(l,m) is a unity normal vector to l.
Call S(c,d) the intersection of l with the perpendicular
dropped on l through P.
Then the distance from P to l is |P,S|.
Now, PS = r.Q and |P,S| = |r| .
PS = r.Q
=> S - P = r. Q
=> S.Q - P.Q = r.Q.Q
=> cl + dm - (al + bm) = r. 1
since S on l holds lc + md = -n
=> -n -al - bm = r
=> la + mb + n = -r
=> |la + mb + n| = |r|
The distance between a point P(a,b)
and a line l with normal equation lx + my + n = 0 is
| la + mb + n | .
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Take lines l and l' with normal equations
lx + my + n = 0 and l'x + m'y + n' = 0 .
A point P(a,b) is on a bisecting line if and only if
distance from P to l = distance from P to l'
<=>
|la + mb + n| = |l'a + m'b + n'|
<=>
la + mb + n = l'a + m'b + n' or la + mb + n = -(l'a + m'b + n')
The two bisecting lines of l and l' with normal equations
lx + my + n = 0 and l'x + m'y + n' = 0 are
lx + my + n = l'x + m'y + n'
and
lx + my + n = -(l'x + m'y + n')
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