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)
(ax + by + c) + t (a'x + b'y + c') = 0Where t is an arbitrary parameter different from 0.
(ax_{o} + by_{o} + c) + t(a'x_{o} + b'y_{o} + c') = 0 + t 0 = 0Conclusion:
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 = -7/3The requested line is -x + 16y - 31 = 0.
ax + by + c = 0If 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.
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.
(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.
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
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. |
Q(ra,rb) is a unit vector <=> Q.Q = 1 <=> ra.ra + rb.rb = 1 <=> r^{2} (a^{2} + b^{2} ) = 1 <=> 1 - 1 r = ------------ or r = ------------ ___________ ____________ V a^{2} + b^{2} V a^{2} + b^{2}Substitute the positive value of r in the equation of l.
ax + by + c l : ----------------- = 0 ____________ V a^{2} + b^{2}This equation is called a normal equation of l.
A normal equation of line ax + by + c = 0 is
ax + by + c ----------------- = 0 ____________ V a^{2} + b^{2} |
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 <=> l^{2} + m^{2} = 1
lx + my + n = 0 is a normal equation of a line <=> l^{2} + m^{2} = 1 |
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 u^{2} + v^{2}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 t^{2} <=> ( 4 + 15 t)^{2} = 4.( 2 + 4 t + 10 t^{2}) <=> ... <=> t = -0.09 of t = -0.47The requested lines are
0.73 x + 1.09 y - 2.09 = 0 -0.41 x + 1.47 y - 2.47 = 0
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 arelx + 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 = 0Find 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 = 0The 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.
_______________________ V (3 + t)^{2} + (4 - t)^{2} = ± 5 <=> ... <=> 2 t^{2} - 2t = 0Since t is not zero, t = 1.
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:
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
S(3,-3) is on r for all t-values <=> (t^{2}-1).3 + t.(-3) + 3 = 0 for all t-values <=> 3 t^{2} - 3 t = 0 for all t-values We see that this is NOT valid for all t-values
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
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,t^{2}) is a variable point of the parabola y = x^{2}. 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. |