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Data defining a straight line in the coordinate system
We can already determine some points with certain characteristics in the coordinate system:
- if the coordinates of the two end-points of a line segment are given, then we can calculate the coordinates of the point of division with an arbitrary ratio and we can also give the length of the line segment;
- if we know the coordinates of the vertices of a triangle, then we can give the coordinates of the centroid.
Our aim now is to give and characterise straight lines. As a result of the investigations we will be able to assign such an equation in two variables to every straight line of the plane that will be satisfied by the coordinates of the points of the straight line and only by these.
A straight line in the plane is unambiguously defined by
- one of its points and its direction,
- two of its points.
The direction of a straight line in the plane can be given in many ways.
Two obvious ways are the following:
Two obvious ways are the following:
- by giving a straight line which is parallel with it,
- by giving a straight line which is perpendicular to it.
Now we are going to investigate how we can express the direction of the straight line with numerical data in the coordinate system.
Direction vector of a straight line
DEFINITION: The direction vector of a straight line is any vector parallel with the straight line and different from the zero vector (Figure 15).
Notation: \latex{\overrightarrow v(v_1; v_2)}.
Notation: \latex{\overrightarrow v(v_1; v_2)}.
The definition implies that if \latex{\overrightarrow v} is a direction vector of the straight line \latex{ e }, then in the case of any real number \latex{\lambda} not equal to \latex{ 0 } \latex{ \lambda\times\overrightarrow v} is also a direction vector of \latex{ e } (Figure 16).
\latex{\overrightarrow v}
\latex{\frac 13 \overrightarrow v}
\latex{-2 \overrightarrow v}
\latex{ 0 }
\latex{ y }
\latex{ x }
\latex{ e }
Figure 16
\latex{\overrightarrow v(v_1;v_2)}
\latex{ y }
\latex{ 0 }
\latex{ x }
\latex{ e }
Figure 15
Example 1
The straight line \latex{ e } passes through the point \latex{P_0(4; –1)}. \latex{\overrightarrow v(2; 4)} is a direction vector of it. Let us calculate the second coordinate of the point \latex{ P } of the straight line \latex{ e } the first coordinate of which is \latex{ 7 }.
Solution
Let us denote the second coordinate of the point \latex{ P } by \latex{ y }. According to the conditions the vector is parallel with \latex{\overrightarrow v}, i.e. there is a real number \latex{\lambda} for which \latex{\lambda \times \overrightarrow v =\overrightarrow{P_0P} } (Figure 17).
As \latex{\overrightarrow{P_0P}= (7 – 4)}\latex{\text i}\latex{ + (y + 1)}\latex{\text j}\latex{ = 3}\latex{\text i}\latex{ + (y + 1)}\latex{\text j}, by writing down the equality for the vectors above with the coordinates \latex{\lambda \times 2 = 3}, and \latex{\lambda \times 4 = y + 1}.
From the first equation \latex{\lambda=\frac3 2} thus the second equation is \latex{\frac{3}{2}\times4=y+1}, which implies \latex{ y = 5 }.
From the first equation \latex{\lambda=\frac3 2} thus the second equation is \latex{\frac{3}{2}\times4=y+1}, which implies \latex{ y = 5 }.
\latex{\overrightarrow{P_0P}}
\latex{P(7; y)}
\latex{P_0(4; -1)}
\latex{\overrightarrow v}
\latex{ x }
\latex{ 0 }
\latex{ -1 }
\latex{ 1 }
\latex{ 2 }
\latex{ 4 }
\latex{ 4 }
\latex{ y }
Figure 17
Example 2
Two points of a straight line are \latex{P_1(–2; 5)} and \latex{P_2(4; 1)}. Let us give a direction vector of this straight line.
Solution
The vector \latex{\overrightarrow{P_1P_2}} is parallel with the straight line and different from the zero vector, therefore it is a suitable direction vector according to the definition (Figure 18).
As \latex{\overrightarrow{P_1P_2}= (4 – (-2))}\latex{\text i}\latex{ + (1 -5 1)}\latex{\text j}\latex{ = 6}\latex{\text i}\latex{ -4}\latex{\text j}, \latex{\overrightarrow v (6; –4)} is a possible solution.
Based on example \latex{ 2 } we can generalise.
If \latex{P1(x_1; y_1)} and \latex{P2(x_2; y_2)}, are two distinct points of the straight line e, then a direction vector of e is \latex{\overrightarrow v(x_2 – x_1; y_2 – y_1)}.
Note: In the case of particular calculations it is worth choosing a direction vector which makes the calculations as easy as possible. Thus for example instead of \latex{\overrightarrow v} we got as a solution in the previous example, \latex{\overrightarrow {v'}=\frac{\overrightarrow v}{2}} is also suitable based on the definition, where \latex{(3; –2)}.
Based on example \latex{ 2 } we can generalise.
If \latex{P1(x_1; y_1)} and \latex{P2(x_2; y_2)}, are two distinct points of the straight line e, then a direction vector of e is \latex{\overrightarrow v(x_2 – x_1; y_2 – y_1)}.
Note: In the case of particular calculations it is worth choosing a direction vector which makes the calculations as easy as possible. Thus for example instead of \latex{\overrightarrow v} we got as a solution in the previous example, \latex{\overrightarrow {v'}=\frac{\overrightarrow v}{2}} is also suitable based on the definition, where \latex{(3; –2)}.
\latex{\overrightarrow{v}}
\latex{P_1(-2; 5)}
\latex{P_2(4; 1)}
\latex{ -2 }
\latex{ 0 }
\latex{ 1 }
\latex{ 4 }
\latex{ 1 }
\latex{ 5 }
\latex{ y }
\latex{ x }
Figure 18
Normal vector of a straight line
DEFINITION: The normal vector of a straight line in the plane is any vector perpendicular to the straight line and different from the zero vector (Figure 19).
Notation: \latex{\overrightarrow n (A; B)}.
Notation: \latex{\overrightarrow n (A; B)}.
\latex{\overrightarrow n(A; B)}
\latex{ 0 }
\latex{ y }
\latex{ x }
\latex{ e }
Figure 19
The definition again implies that if \latex{\overrightarrow n} is a normal vector of the straight line \latex{ e }, then for any real number \latex{\lambda} not equal to \latex{0, \lambda \times \overrightarrow n} is also a normal vector of e (Figure 20)
\latex{2\times\overrightarrow n}
\latex{\overrightarrow n}
\latex{-3\times\overrightarrow n}
\latex{ 0 }
\latex{ y }
\latex{ x }
\latex{ e }
Figure 20
Example 3
The straight line e passes through the point \latex{P_0(3; 4)}, one of its normal vectors is \latex{\overrightarrow n(–2; 5)}. Let us calculate the first coordinate of the point \latex{ P } of the straight line the second coordinate of which is \latex{ 7 }.
Solution
During the solution we are using a theorem we have already proven:
The dot product of two vectors is \latex{ 0 } if and only if the two vectors are perpendicular to each other.
The dot product of two vectors is \latex{ 0 } if and only if the two vectors are perpendicular to each other.
The point \latex{ P } lies on e if and only if the vectors \latex{\overrightarrow {PP_0}} and \latex{\overrightarrow n} are perpendicular to each other. If \latex{ x } denotes the first coordinate of the point \latex{ P }, then
\latex{\overrightarrow{P_0P}=(x-3)}\latex{\text{i}}\latex{+(7-4)}\latex{\text{j}}\latex{=(x-3)}\latex{\text{i}}\latex{+3}\latex{\text{j}}.
By writing down the dot product of the two vectors in terms of the coordinates:
\latex{\overrightarrow{P_0P}\times\overrightarrow n=(x-3)\times(-2)+3\times5=-2x+21=0},
which implies \latex{x=\frac{21}{2}=10.5}.
Example 4
Let us calculate the area of the triangle defined by the points \latex{A(-2; 2)}, \latex{B(3; 3)}, \latex{C(8; -2)}.
Solution
If \latex{\overrightarrow n(A; B)} is a normal vector of the straight line, then the vectors \latex{\overrightarrow n} and \latex{\overrightarrow{P_1P_2}} are perpendicular to each other. However it implies that their dot product is \latex{ 0 }. Since
\latex{\overrightarrow{P_1P_2}=(-2-3)}\latex{\text{i}}\latex{+(6-(-4))}\latex{\text j}\latex{=-5}\latex{\text i}\latex{+10}\latex{\text j},
by writing down the dot product in terms of the coordinates
\latex{\overrightarrow{P_1P_2}\times \overrightarrow n = –5A + 10B = 0}.
It can be seen that one of \latex{ A } and \latex{ B } can freely be selected; however the other one is then unambiguously determined from this selection. By paying attention to the symmetry, for example the selection \latex{ A = 10 }, \latex{ B = 5 } is suitable. So a normal vector of the straight line considered is \latex{\overrightarrow n(10; 5)}.
Note: In order to simplify the particular calculations it is worth selecting the normal vector out of the possible normal vectors which make the coordinate calculations as easy as possible. Thus \latex{\overrightarrow n (2; 1)} is also a suitable normal vector in example \latex{ 4 }, and it can be dealt with more easily while calculating.
A direction vector of the straight line in example \latex{ 4 } is \latex{\overrightarrow{P_1P_2}= –5}\latex{\text i}\latex{ + 10}\latex{\text j}, a normal vector can be \latex{\overrightarrow n = 10}\latex{\text i}\latex{ + 5}\latex{\text j}. It can be realised that the magnitudes of these two vectors perpendicular to each other are equal, and one of them can be derived from the other one by replacing the coordinates and changing one of them to its negative value.
The remark we made in connection with the specific example also holds in general:
If a direction vector of the straight line \latex{ e } is \latex{\overrightarrow v (v_1; v_2)}, then the vectors \latex{\overrightarrow n (v_2; –v_1)} and \latex{\overrightarrow n (-v_2; v_1)} are the normal vectors of \latex{ e }.
Angle of inclination (or azimuth angle)
and gradient of a straight line
and gradient of a straight line
In the planar coordinate system we have the chance to give the direction of a straight line with the help of the signed angle of inclination included between the straight line and one of the axes or with the help of a trigonometric function of this angle.
DEFINITION: In the planar coordinate system the angle of inclination (or azimuth angle) of a straight line is the signed angle included between the straight line and the positive ray (positive direction) of the \latex{ x }-axis (Figure 21).
For the angle of inclination \latex{\alpha}, \latex{-\frac{\pi}{2}\lt \alpha\leq\frac{\pi}{2}}, and \latex{\alpha} is positive or negative depending whether the \latex{ x }-axis needs to be rotated to the positive or the negative direction about the origin by the angle from the interval allowed so that its rotated image will be parallel with the straight line.
The angle of inclination (or azimuth angle) of the \latex{ y }-axis is \latex{\frac{\pi}{2}}.
The angle of inclination (or azimuth angle) of the \latex{ y }-axis is \latex{\frac{\pi}{2}}.
In the case of specific calculations the direction of the straight line is often characterised with the tangent of the angle of inclination (if exists) instead of the angle of inclination.
\latex{\alpha\gt0}
\latex{\beta\lt0}
\latex{\beta}
\latex{\alpha}
\latex{ 0 }
\latex{ x }
\latex{ y }
\latex{ e }
\latex{ f }
Figure 21
DEFINITION: The tangent of the angle of inclination of a straight line (if it exists) is called the gradient of the straight line.
The gradient of the straight line with an angle of inclination of \latex{\alpha} is \latex{m = \tan\alpha}.
The definition of the angle of inclination implies that the gradient of straight lines parallel with the \latex{ y }-axis does not exist, and the gradient of straight lines parallel with the \latex{ x }-axis is \latex{ 0 }.
Example 5
Let us calculate the gradient and the angle of inclination of the straight line defined by the points \latex{P_1(–5; 1)} and \latex{P_2(1; 6)}.
\latex{\alpha}
\latex{\alpha}
\latex{1-(-5)=6}
\latex{6-1=5}
\latex{ 6 }
\latex{ 1 }
\latex{ 0 }
\latex{ -5 }
\latex{ P_{1} }
\latex{ P_{2} }
\latex{ Q }
Figure 22
Solution
In Figure 22 in the right-angled triangle \latex{P_1QP_2} \latex{\alpha = P_2P_1Q\sphericalangle} is the angle of inclination of the straight line.
As \latex{P_2Q = 6 – 1 = 5} and \latex{P_1Q = 1 – (–5) = 6}, \latex{m = \tan\alpha =\frac{P_2Q}{P_1Q}=\frac{5}{6}} and thus \latex{\alpha\approx 39.81^{\circ}}.
The method applied to determine the gradient can be applied in the case of any straight line given by two of its points and not parallel with the y-axis.
The method applied to determine the gradient can be applied in the case of any straight line given by two of its points and not parallel with the y-axis.
\latex{P_1(x_1; y_1)}
\latex{P_2(x_2; y_2)}
\latex{x_2-x_1}
\latex{y_2-y_1}
\latex{ y }
\latex{ 0 }
\latex{ x }
\latex{ e }
Figure 23
If \latex{P_1(x_1; y_1)} and \latex{P_2(x_2; y_2)} are two distinct points of the straight line e, where \latex{x1 \neq x2}, then the gradient of the straight line e is (Figure 23)
\latex{m=\frac{y_2-y_1}{x_2-x_1}}.
The above imply that the gradient of the straight line can also be expressed in terms of the coordinates of the direction vector and the normal vector.
If \latex{\overrightarrow v(v_1; v_2)} is a direction vector and \latex{\overrightarrow n (A; B)} is a normal vector of e; and \latex{v_1 \neq 0}, and \latex{B\neq 0}, then the gradient of the straight line e is (Figure 24)
\latex{m=\frac{\nu_2}{\nu_1}} and \latex{m=-\frac{A}{B}}
\latex{\overrightarrow n(A; B)}
\latex{\alpha}
\latex{\alpha}
\latex{\overrightarrow v(v_1; v_2)}
\latex{ 0 }
\latex{ x }
\latex{ y }
\latex{ e }
Figure 24
A consequence of the relations above:
If the gradient of the straight line \latex{ e } is \latex{ m }, then \latex{\overrightarrow v (1; m)} is a direction vector and \latex{\overrightarrow n (m; –1)} (or \latex{\overrightarrow n (-m; 1)}) is a normal vector of \latex{ e }.
Example 6
In the coordinate system \latex{\overrightarrow v (–3; 1)}. is a direction vector of the straight line \latex{ e }. Let us give a normal vector, the gradient and the angle of inclination of the straight line.
Solution
Based on the relations above \latex{\overrightarrow{n} (1; 3)} is a normal vector of \latex{ e }, its gradient is \latex{m =-\frac{1}{3}} its angle of inclination is \latex{\alpha \approx –18.43^{\circ}}.
Conditions for two straight lines to be parallel or perpendicular
- Let us assume that the straight lines \latex{ e_1 } and \latex{ e_2 } are parallel (Figure 25) and that their gradient exists (their angle of inclination is different from \latex{-\frac{\pi}{2}}).
Let the angle of inclination of \latex{e_1} be \latex{\alpha1}, let its gradient be \latex{m_1}, let a direction vector be \latex{\overrightarrow {v_1}}, and let a normal vector be \latex{\overrightarrow{n_1}}. Let the corresponding direction characteristics of \latex{\overrightarrow{e_2}} be \latex{\overrightarrow{\alpha_1}}, \latex{\overrightarrow{m_2}}, \latex{\overrightarrow{v_2}}, \latex{\overrightarrow{n_2}}.
\latex{\overrightarrow {v_1}}
\latex{\overrightarrow {v_2}}
\latex{\overrightarrow {n_2}}
\latex{\overrightarrow {n_1}}
\latex{\alpha_1}
\latex{\alpha_2}
\latex{ 0 }
\latex{ x }
\latex{ e_1 }
\latex{ y }
\latex{ e_2 }
Figure 25
It can be seen that in this case
\latex{\alpha_1 = \alpha_2}, and thus \latex{m_1 = m_2},
and there are such real numbers \latex{\lambda} and m different from 0 that
\latex{\overrightarrow{v_1} = \lambda \times \overrightarrow {v_2}} and \latex{\overrightarrow {n_1} = \mu \times \overrightarrow{n_2}}.
It can easily be proven that if any of the conditions given for the direction characteristics above is satisfied, then the two straight lines are parallel, so any of these four conditions is necessary and sufficient for straight lines not parallel with the \latex{ y }-axis to be parallel.
- Let us assume that the straight lines \latex{ e_1 } and \latex{ e_2 } are perpendicular to each other (Figure 26) and are not parallel with the coordinate axes.
If the direction characteristics of \latex{ e_1 } are \latex{\alpha_1}, \latex{m_1}, \latex{\overrightarrow{v_1}(v_1 ’; v_2 ’)}, \latex{\overrightarrow{n_1}(A_1; B_1)}, and the direction characteristics of the straight line \latex{e_2} are \latex{a_2}, \latex{m_2}, \latex{\overrightarrow{v_2}(v_1 ”; v_2 ”)}, \latex{\overrightarrow{n_2}(A_2; B_2)}, then the difference of the angles of inclination is \latex{\frac{\pi}{2}}, and the dot products of the two direction vectors and the two normal vectors are both \latex{ 0 }, i.e.
\latex{\overrightarrow{v_1} \times \overrightarrow{v_2} = 0} and \latex{\overrightarrow{n_1} \times \overrightarrow{n_2} = 0}.
By writing down the dot product of the direction vectors with the coordinates
\latex{v_1'\times v_1''+v_2'\times v_2''=0}, which implies \latex{\frac{v_2'}{v_1'}=-\frac{v_1''}{v_2''}=-\frac{1}{\frac{v_2''}{v_1''}}}.
\latex{\overrightarrow {v_1}}
\latex{\overrightarrow {v_2}}
\latex{\alpha_1}
\latex{\alpha_2}
\latex{ 0 }
\latex{ x }
\latex{ y }
\latex{ e_1 }
\latex{ e_2 }
Figure 26
By using that the gradient of a straight line (if exists) is obtained from a direction vector of the straight line as the corresponding quotient of the coordinates, we get the following:
\latex{m_1=-\frac{1}{m_2}},or in another form \latex{m_1\times m_2=-1}.
The train of thought can be reversed; therefore two straight lines not parallel with the coordinate axes are perpendicular to each other if and only if the product of their gradients is \latex{ –1 }.
In specific calculation exercises we will often use that the direction vector of a straight line given is the normal vector of any straight line perpendicular to it, while the normal vector of a straight line given is the direction vector of any straight line perpendicular to it. In this case it means that \latex{\overrightarrow {n_2}} is a direction vector of \latex{e_1}, \latex{\overrightarrow{v_2}} is a normal vector of it, while \latex{\overrightarrow{v_2}} is a direction vector of \latex{ e_2 }, and \latex{\overrightarrow{v_1}} is a normal vector of it.
Example 7
The straight lines \latex{ f } and \latex{ g } are perpendicular to each other. The vector \latex{\overrightarrow n (2; 1)} is a normal vector of the straight line \latex{ f }. Let us calculate the gradient of the straight line g, and let us give a direction vector and a normal vector for it.
Solution
The given normal vector of the straight line \latex{ f } is a direction vector of \latex{g: \overrightarrow{v_g}(2; 1)}. It implies that the gradient of \latex{ g } is \latex{m_g= \frac{1}{2}}, and \latex{\overrightarrow {n_g}(–1; 2)} is a normal vector of it.
Exercises
{{exercise_number}}. A point on a straight line is \latex{P_0(0; 2)}, a direction vector is \latex{\overrightarrow v (3; 1)}. Calculate the first coordinate of the point P on the straight line the second coordinate of which is
- \latex{0};
- \latex{4};
- \latex{-1};
- \latex{7};
- \latex{\frac{2}{3}};
- \latex{\sqrt 2}.
{{exercise_number}}. A point on a straight line is \latex{P_0(–1; 3)}, a normal vector is \latex{\overrightarrow n(–2; 5)}. Give the second coordinate of the point \latex{ P } lying on the straight line if its first coordinate is
- \latex{0};
- \latex{4};
- \latex{-1};
- \latex{7};
- \latex{\frac{2}{3}};
- \latex{\sqrt 2}.
{{exercise_number}}. Does the point \latex{ P } coincide with the straight line passing through the point \latex{P_0(4; 3)} and with the direction vector \latex{\overrightarrow v (–2; 5)}, if
- \latex{P(0; 0)};
- \latex{P(2; 8)};
- \latex{P(8; -7)};
- \latex{P(-3; 1)};
- \latex{P(-2; 6)}?
{{exercise_number}}. Does the point \latex{ P } coincide the straight line passing through the origin and with the normal vector \latex{\overrightarrow n (3; 2)}, if
- \latex{P(2; -3)};
- \latex{P(1; 2)};
- \latex{P(-1; -7)}:
- \latex{P(4; 6)};
- \latex{P\left(-1; \frac{3}{2}\right)}?
{{exercise_number}}. Give a direction vector, a normal vector, the gradient and the angle of inclination of the straight line \latex{P_1P_2}, if
- \latex{P_1(0; 0), \;\;P_2(3; -2)};
- \latex{P_1(1; 3), \;\;P_2(2; 4)};
- \latex{P_1(-4; 7), \;\;P_2(6; -5)}.
{{exercise_number}}. A normal vector of a straight line is
- \latex{\overrightarrow n(0; 2)};
- \latex{\overrightarrow n(1; -2)};
- \latex{\overrightarrow n(3; 5)};
- \latex{\overrightarrow n(-4; -10)}.
Give a direction vector, the gradient and the angle of inclination of the straight lines in each case.
{{exercise_number}}. The straight lines \latex{ e } and \latex{ f } are perpendicular to each other. The gradient of \latex{ e } is
- \latex{1};
- \latex{5};
- \latex{\frac{1}{2}};
- \latex{-3};
- \latex{\sqrt5};
- \latex{0}.
Calculate the gradient and the angle of inclination of the straight line \latex{ f }.
{{exercise_number}}. A direction vector of the straight line \latex{ e } is \latex{\overrightarrow{v_e} (1; 3)}, a direction vector of the straight line \latex{ f } is \latex{\overrightarrow{v_f} (-6; y)}. Calculate \latex{ y } if we know that \latex{ e } and \latex{ f }
- are parallel with each other;
- are perpendicular to each other.