Lines in a plane - Orthogonality; Distances

Variable point on a line

Consider the line d with equation y = m x + q . Let t be a variable real number. Point P(t, m.t+q) is on the line d. P is a variable point of the line. The variable t is called a 'parameter'.

Example: P(t, 3t-1 ) is a variable point on the line y = 3x-1.

Application:
Find the intersection on the line d with equation y = 3x-1 and the line d' with equation 4 x+ 3y -1=0.

Start with a variable point P(t, 3t-1 ) on d.
P is on d' if and only if 4 t + 3(3t-1) - 1 = 0 <=> t = 4/13.
The intersection point is P(4/13 , -1/13)

Concurrent lines

Three or more lines are said to be concurrent if they intersect at a single point.

Equation of a line through the intersection point of two given lines

We start with two intersecting lines ax + by + c = 0 en a'x + b'y + c' = 0.
Say S(xo, yo) is the intersection point. Then we have axo + byo + c = 0 and a'xo + b'yo + c' = 0.
Now we consider line d with equation 
 
   (ax + by + c) + t (a'x + b'y + c') = 0
Where t is an arbitrary parameter different from 0.
The point S is on line d because 
 
(axo + byo + c) + t(a'xo + b'yo + c') = 0 + t 0 = 0
Conclusion: 
 
  For each t, different from 0, the line

    (ax + by + c) + t (a'x + b'y + c') = 0

  is a variable line through the intersection point of

  ax + by + c = 0  and   a'x + b'y + c' = 0 .

  t is a parameter.

Example:

Calculate the equation of the line d that connects point P (1.2) with the intersection point of the lines 2x+3y-1=0 and x-y+4=0.

A variable line d through the intersection of 2x+3y-1=0 and x-y+4=0 has an equation (2x+3y-1) + t(x-y+4) = 0.
Now we can calculate the appropriate value of t such that the line d contains the point P.
The condition is : 

 
   (2.1+3.2-1) + t(1-2+4) = 0
<=>
    t = -7/3
The requested line is -x + 16y - 31 = 0.

Orthogonal lines

Orthonormal-axes

In all that follows we assume: We call this : 'two orthonormal axes x and y'.

Equation and slope of a line

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 .
Lines with the same slope are parallel.

Remark: ax + by = 0 is the equation of a line parallel to l and containing the origin O.

Examples:

Line y = 3x + 4 is parallel to the line 6x - 2y + 7 = 0

Line 5x - 3y +3 = 0 is parallel to the line 3y - 5x -14 = 0.

Find t such that line 2x + (t-1) y + 4 = 0 is parallel to the line x + y -12= 0.
Answ: t=3.

Direction vector of a line

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).
(b,-a) are the coordinates of a direction vector of the line ax + by + c = 0

Examples:

(3,5) are the coordinates of a direction vector of the line 5x - 3y +3 = 0.

(1,0) are the coordinates of a direction vector of the x-axis.

(0,1) are the coordinates of a direction vector of the y-axis.

(0,-7) are the coordinates of a direction vector of the y-axis.

Orthogonal lines formula

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
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 .

Examples:

3x + 5y - 12 = 0 and 5x - 3y - 55 are orthogonal.

y - 4 = 0 and 3 x +16 = 0 are orthogonal.

Find t such that 2x + (t-1) y + 4 = 0 and x + y -12= 0 are orthogonal.
Ans: t = -1 

 
Find the line k through the intersection of
    line  a :  2x + 3y -7 = 0
and line  b :  x + y = 0
such that  k is orthogonal to line  c :  x - 3y + 10 = 0

We take a variable line k through the intersection of a and b. 

 
   (2x + 3y -7) + t (x + y) = 0
<=>
   (2+t)x + (3+t)y -7 = 0

   k is orthogonal to line  c
<=>
     1(2+t) + 3(3+t) = 0
<=>
     11 + 4t = 0
<=>
     t = -11/4

  Line  k is

   (2-11/4)x + (3-11/4)y - 7 = 0
<=>
   3x - y + 28 = 0

Orthogonal lines and slope

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.

Distance from a point to a line

Normal vector to a line

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.

Normal equation of a line

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-value such that the normal vector Q(ra,rb) is a unit vector. 
 
        Q(ra,rb) is a unit 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.

A normal equation of line ax + by + c = 0 is 
 
               ax +  by + c
             ----------------- = 0
                ____________
               V a2  + b2

Examples:

A normal equation of the line 3 x + 4 y -10 = 0 is 3/5 x + 4/5 y - 2 = 0

A normal equation of the y-axis is x=0

A normal equation of the line y = x is (x - y)/sqrt(2) = 0

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

 
            lx + my + n = 0 is a normal equation of a line

                        <=>

                      l2  + m2  = 1

Distance between a point and a line

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 from point P(a,b) to the line ux + vy + w = 0 is 
 
               u a +  v b + w
           | ----------------- |
                ____________
               V u2  + v2
Mind the absolute value signs

Example 1:

We calculate the distance from P(1,-1) to the line 3x + 4y -1 = 0.

The normal equation of the line is ( 3x + 4y -1 )/5 = 0.

Then the distance is | (3.1 + 4.(-1) -1)/5 | = 2/5

Example 2

The distance from the origin to the line x + 2y -4 = 0 is 4/sqrt(5).

Example 3:

Find the lines through the intersection of x + y - 2 = 0 and 3x - y + 1 = 0 that are tangent to the circle with center M(5,1) and with radius 2.

Procedure: Take a variable line v through the intersection of x + y - 2 = 0 and 3x - y + 1 = 0. This line v is tangent to the circle if and only if the distance from M to v is equal to 2.

The variable line v has a equation with parameter t 

 
    (x + y - 2) + t (3x - y + 1) = 0
<=>
    (1 + 3t) x + (1 - t) y + (-2 + t) = 0



    |M , v| = 2
<=>
      (1 + 3t) 5 + (1 - t)  + (-2 + t)
   | ---------------------------------- | = 2
        _________________________
       V (1 + 3t)2 + (1 - t)2
<=>
                    ____________________
   | 4 + 15 t| = 2 V 2 + 4 t + 10 t2
<=>
   ( 4 + 15 t)2 = 4.( 2 + 4 t + 10 t2)
<=>
    ...
<=>
    t = -0.09  of t = -0.47
The requested lines are 
 
  0.73 x + 1.09 y - 2.09 = 0

 -0.41 x + 1.47 y - 2.47 = 0

Bisecting lines

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')

Example 1

Take the lines k : 3x +4y +10 = 0 and l : x + y + 2 = 0 .
The normal equations of the lines are (3/5)x +(4/5)y +2 = 0 and (x + y + 2)/sqrt(2) = 0
The distance from P(0,1) to k is 14/5.
The two bisecting lines of k and l are
(3/5)x +(4/5)y +2 = (x + y + 2)/sqrt(2)
(3/5)x +(4/5)y +2 = - (x + y + 2)/sqrt(2)

Example 2
Given: the lines 
 
   k :   3 x + 4 y - 7 = 0

   b :   x - y = 0
Find the line m such that b is a bisecting line of k and m.

The line m is a line through the intersection point of k and b.
The equation of m has the form 

 
  ( 3 x + 4 y - 7) + t( x - y) = 0
<=>
  (3 + t) x + (4 - t) y - 7 = 0
The parameter t is different from zero. We'll calculate the t-value such that b is a bisecting line of k and m. 
 
 The  normal equation of k is

(3 x + 4 y - 7)/5 = 0

 The  normal equation of m is

  (3 + t) x + (4 - t) y - 7
  -------------------------- = 0
   ______________________
  V (3 + t)2 + (4 - t)2

 The bisecting lines of k and m are

  (3 + t) x + (4 - t) y - 7
  -------------------------- = ± (3 x + 4 y - 7)/5
   ______________________
  V (3 + t)2 + (4 - t)2
We search the value of t such that this line coincides with x - y = 0.
In any case, the constant term must disappear. The condition is : 
 
   _______________________
  V (3 + t)2 + (4 - t)2 =  ± 5

<=>
   ...
<=>
   2 t2 - 2t = 0
Since t is not zero, t = 1.
The requested equation of m is 4x + 3 y - 7 = 0.

Direction vector - collinear points - concurrent lines

These subjects are covered on this page., but it requires knowledge of the theory of matrices, determinants and systems of equations.

Does a variable line goes through a fixed point ?

A variable tangent line to a parabola is a variable line, but there is no fixed point S so that S always is on that tangent line.

It can happen that a variable line is given, and that we should investigate whether or not that line contains a fixed point.

This can be done in a two step method:

  1. We take two different positions r1 and r2 of the variable line r. We calculate the intersection point S of r1 and r2.
    If there exist a fixed point, then S is that point.
  2. We will examine whether S is actually the fixed point of r, by examining whether S is on r independently of the position of r.
We now give three examples of such research.

Example 1

We examine whether the variable line r with equation (2+t)x + (3-t)y -8 + t = 0
contains a fixed point. The position of r is different for each t-value.

  1. We take two, well chosen, values of t. We obtain two positions r1 and r2 of the line.
    Take t = -2 ; the line r1 is 5 y = 10 <=> y = 2.
    Take t = 3 ; the line r2 is 5x = 5 <=> x = 1.
    The intersection point is S(1,2)
  2. We examine whether S is on r independently of the position of r. 
     
   S(1,2) is on r for all t-values
<=>
   (2+t)(1) + (3-t) 2 - 8 + t = 0   for all t-values

   We see that this is valid for all t-values
Conclusion: S(1,2) is a fixed point of the variable line r.

Example 2

We examine whether the variable line r with equation
(t2-1)x + t y + 3 = 0
contains a fixed point. The position of r is different for each t-value.

  1. We take two, well chosen, values of t. We obtain two positions r1 and r2 of the line.
    Take t = 1 ; the line r1 is y = -3
    Take t = 0 ; the line r2 is -x = -3
    The intersection point is S(3, -3)
  2. We examine whether S is on r independently of the position of r. 
     
     S(3,-3) is on r for all t-values
<=>
    (t2-1).3 + t.(-3) + 3 = 0 for all t-values
<=>
     3 t2 - 3 t = 0  for all t-values

   We see that this is NOT valid for all t-values
Conclusion: the variable line has no fixed point.

Example 3

We examine whether the variable line r with equation
(3 sin(t)-2)x - 3 cos(t) y + 2 cos(t) = 0
contains a fixed point. The position of r is different for each t-value.

  1. We take two, well chosen, values of t. We obtain two positions r1 and r2 of the line.
    Take t = pi/2 ; the line r1 is x = 0
    Take t = 0 ; the line r2 is -3 y + 2 = 0
    The intersection point is S(0, 2/3)
  2. We examine whether S is on r independently of the position of r. 
     
    S(0, 2/3)  is on r for all t-values
<=>
    (3 sin(t)-2).0 - 3 cos(t)(2/3) + 2 cos(t) = 0  for all t-values
<=>
     -2 cos(t) + 2 cos(t) = 0   for all t-values

   We see that this is valid for all t-values
Conclusion: S(0, 2/3) is a fixed point of the variable line r.

Exercises

Find the line d through the intersection point of the lines 2x+y-3=0 and x+y-2=0 and perpendicular to the line 3x-y+4=0.

The line d has equation x + y + 1 = 0 and point P has coordinates (1,2). Find the points Q on d such that |PQ|=4.

The lines d and d' have respectively an equation 3x + 4 y + 1 = 0 and x - y - 2 = 0. Find the points P on d' such that the distance from P to the line d is 7.

The following exercises are also based the knowledge of the properties of 'dot product'. ( see vectors )

The angle between two lines d and d' is equal to the sharp angle between a direction vector of d and a direction vector of d'.
Find the angle ( in degrees) between two lines d and d' with equation 3x + 4 y + 1 = 0 and x - y - 2 = 0.

Two lines d and d' have as equation 3x + 4 y + 1 = 0 and x - y - 2 = 0.
Say b en b' are the bisecting lines of d and d'. Find the angle (in degrees) between the lines d and b and between the lines d and b'.

Point P(t,t2) is a variable point of the parabola y = x2.
Consider the line PQ with Q(6,0). Find t such that the angle, between PQ and the line y=0, is 45 degrees.

A regular pentagon has its center in point O(0,0) and its side [AB] is on the line z with equation 3x + 4y -12 = 0.
Calculate the area of the pentagon.



Topics and Problems

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