cadaVR anatomy by Mozaik Education

The intersection point, the distance and
the angle of inclination of two straight lines

The intersection point of two straight lines

Example 1

Let us calculate the coordinates of the intersection point of the straight line passing through the point \latex{A(2;-1)} with a normal vector \latex{\overrightarrow{n}(2;3)} and the straight line passing through the point \latex{B(3;2)} with the gradient \latex{\frac{1}{4}}.

Solution

Let us first write down the equations of the two straight lines. Let \latex{ a }
denote the straight line passing through the point \latex{ A }, and let \latex{ b } denote
the straight line passing through the point \latex{ B }.
The equation of \latex{ a } based on the general form \latex{(A=2,B=3,x_{0}=2,y_{0}=-1 ):}

\latex{2x+3y=1.}

And the equation of \latex{ b } from the gradient form \latex{(m=\frac{1}{4},x_{0}=3,y_{0}=2 ):}

\latex{y-2=\frac{1}{4}\times (x-3),} which implies \latex{x-4y=-5.}

The coordinates of the intersection point of the two straight lines satisfy the equations of both straight lines, therefore these are the solutions of the linear system of equations in two variables consisting of the equations of the two straight lines. So we have to solve the simultaneous equations below to find the coordinates of the intersection point:

\latex{\begin{rcases}2x+3y=1 \\ x-4y=-5\end{rcases}.}

For example the method of equal coefficients can be applied to solve
the simultaneous equations. By subtracting two times the second
equation from the first equation we obtain

\latex{11y=11,}

hence \latex{y=1}. By substituting it back into the second equation we obtain \latex{x=-1.}
So the intersection point of the straight lines a and b is \latex{M(-1;1)} (Figure 35).
Note: It is possible that in the case of two straight lines the system of equations consisting of their equations does not have a solution. It happens exactly when the two straight lines are parallel but do not coincide. Such equations are for example the straight lines \latex{e:x-2y=5} and \latex{f:-2x+4y=-2} (Figure 36).

That they are parallel can be realised even before writing down the
system of equations consisting of the equations of the two straight lines,
if we find one of the direction characteristics of the straight lines from
the equations. In the case of this example the gradient of both straight lines is \latex{\frac{1}{2}.}

The method applied in example \latex{ 1 } holds in general:

The coordinates of the intersection point of two coplanar and intersecting straight lines are the solutions of the system of equations consisting of the equations of the two straight lines.

The distance between a point and a straight line

Example 2

Let us calculate the distance between the straight line e with the equation \latex{x-2y=4} and the point \latex{P(3;6)}.

Solution

The distance between a point and a straight line is the distance between
the given point and the base point of the line segment through the point
perpendicular to the given straight line (Figure 37).
To determine the distance \latex{ PT } we need the coordinates of the point \latex{ T }.
These are obtained as the solutions of the system of equations consisting of the equation of the straight line \latex{ f } passing through \latex{ P } and perpendicular to e and the equation of the straight line \latex{ e }. Thus the steps of the solution are:

setting up the equation of \latex{ f } ;
solving the system of equations consisting of the equations of \latex{ e } and \latex{ f }; the coordinates of \latex{ T } are obtained;
determining the distance \latex{ PT }.

\latex{ e } is perpendicular to \latex{ f }, so a normal vector of \latex{ e } is a direction vector of \latex{ f }. A normal vector of \latex{ e } can be found from its equation: \latex{\overrightarrow{n_{e} }(1;-2).} So a direction vector of \latex{ f } is \latex{\overrightarrow{v_{f} }(1;-2).}

So a point (P) and a direction vector \latex{(\overrightarrow{v_{f} } )} of \latex{ f } are given, thus by using the direction vector form of the straight line the equation can be set up
\latex{(v_{1}=1,v_{2}=-2,x_{0}=3,y_{0}=6 ):}

\latex{-2x-y=(-2)\times 3-1\times 6,}

which implies after subtracting and multiplying by \latex{ –1 }

\latex{2x+y=12.}

The simultaneous equations to solve:

\latex{\begin{rcases}x-2y=4 \\ 2x+y=12\end{rcases};}

By subtracting twice the first equation from the second equation
\latex{5y=4} is obtained, which implies \latex{y=\frac{4}{5}.} By substituting it into the first equation

\latex{x=2\times \frac{4}{5}+4=\frac{28}{5}=5\frac{3}{5}.}

So the intersection point of \latex{ e } and \latex{ f } is found to be \latex{ T } \latex{(\frac{28}{5};\frac{4}{5} )}.

\latex{PT=\sqrt{(3-\frac{28}{5} )^{2}+(6-\frac{4}{5} )^{2} }=\sqrt{(-\frac{13}{5}^{2}+(\frac{26}{5} )^{2} }=}\latex{=\sqrt{\frac{845}{25} }=\sqrt{\frac{169}{5} }=\frac{13}{\sqrt{5} }\approx 5.81.}

So the distance between the point \latex{ P } and the straight line \latex{ e } is \latex{\approx 5.81.}
The method used in the solution of example \latex{ 2 } can be applied for determining the distance between any straight line and a point not lying on it.
Note: The distance between a straight line and any point lying on it is \latex{ 0 }.

The distance between two parallel straight lines

Example 3

Let us calculate the distance between the straight lines with the equations

\latex{3x+2y=12} and \latex{3x+2y=-6.}

Solution I (draft)

As the distance of two parallel straight lines is the distance between an arbitrary point of one of the straight lines and the other straight line, let us take the point of the straight line \latex{ e } with the equation \latex{3x+2y=12} lying on the \latex{ x }-axis. As the second coordinate of this is \latex{ 0 }, the equation of the straight line implies that \latex{x=4}. (Figure 38)

So our task is to determine the distance between the point \latex{P(4;0)} and the straight line \latex{ f } with the equation \latex{3x+2y=-6.} It can be done with the solution method of example \latex{ 2 }.

Solution II

The distance of two parallel straight lines can also be determined by taking a straight line perpendicular to both of them and by determining the distance of the two intersection points obtained. It is practical to take a perpendicular straight line the equation of which is simple so we can relatively easily calculate the intersection points with the parallel
straight lines.
According to the above let us consider the straight line \latex{ g } perpendicularly intersecting the straight lines \latex{ e } and \latex{ f } and passing through the origin (Figure 39). The equation of \latex{ g } has a form of \latex{y=mx,}and the gradient \latex{ m } can be determined from the equation of \latex{ e } or \latex{ f }. Thus the steps of the solution are:

setting up the equation of \latex{ g } (determining \latex{ m });
determining the coordinates of \latex{e\cap g=M_{1}} and \latex{f\cap g=M_{2};}
calculating the distance \latex{M_{1}M_{2}.}

\latex{\overrightarrow{n}(3;2)} is a normal vector of the straight lines \latex{ e } and \latex{ f }. It is a direction vector of \latex{ g }: \latex{\overrightarrow{v_{g} }(3;2).} Thus the gradient of \latex{ g } is \latex{m=\frac{2}{3} (v_{1}=3;v_{2}=2, m=\frac{v_{2} }{v_{1} } ) .} So the equation of \latex{ g } is \latex{y=\frac{2}{3}\times x.}

We have to solve the simultaneous equations

\latex{\begin{rcases}3x+2y=12 \\ y=\frac{2}{3}\times x \end{rcases}}

to find the coordinates of \latex{M_{1}.} By applying the substitution method we
obtain the following:

\latex{3x+\frac{4}{3}\times x=12,}
\latex{x=\frac{36}{13}.}

By substituting it into the second equation we get \latex{y=\frac{24}{13}.} Therefore

\latex{M_{1}(\frac{36}{13};\frac{24}{13} ).}

The coordinates of \latex{M_{2}} are obtained similarly:

\latex{\begin{rcases}3x+2y=-6 \\ y=\frac{2}{3}\times x \end{rcases}.}

By substituting y in the first equation

\latex{3x+\frac{4}{3}\times x=-6,}
\latex{x=-\frac{18}{13}.}

From the second equation \latex{y=-\frac{12}{13}.} Therefore

\latex{M_{2}(-\frac{18}{13};-\frac{12}{13} ).}

By substituting the corresponding coordinates into the distance formula

\latex{M_{1}M_{2}=\sqrt{(\frac{36}{13}-(-\frac{18}{13} ) )^{2}+(\frac{24}{13}-(-\frac{12}{13} ) )^{2} }=\sqrt{(\frac{54}{13} )^{2}+(\frac{36}{13} )^{2} }=}

\latex{=\frac{\sqrt{4212} }{13}=\frac{18\times \sqrt{13} }{13}\approx 4.99.}

So the distance of the straight lines \latex{ e } and \latex{ f } is \latex{\approx 4.99\approx 5.}
Both solution methods of example \latex{ 3 } can be applied to determine the distance of parallel straight lines.

The angle of inclination of two straight lines
(higher level courseware)

Two coplanar intersecting straight lines generate four angular domains. The two opposite ones are congruent (vertical angles) respectively. The smaller (not larger) one of the two adjacent angles generated is called the angle of inclination of the two straight lines
(Figure 40).

Example 4

Let us determine the angle of inclination of the straight lines with the equations \latex{2x-3y=6} and \latex{4x+y=8.}

Solution

The angle included between two vectors not equal to the zero vector can be determined with the help of the dot product of the two vectors. The angle of inclination of two straight lines or its supplementary angle is equal to the angle of inclination of the direction vectors of the straight lines, thus we have to calculate the angle included between these direction vectors to determine the angle of inclination of the two straight lines.
The direction vectors of the straight lines can be found from the equations:
a direction vector of the straight line e with the equation \latex{2x-3y=6} is \latex{\overrightarrow{v_{e} }(3;2);}

In terms of the cosine of the angle \latex{\varphi} defined by the two direction vectors from the two expressions of the dot product

\latex{\cos \varphi= \frac{\overrightarrow{v_{e}\ }\times \overrightarrow{v_{f} } }{|\overrightarrow{v_{e}| }\times |\overrightarrow{v_{f}| } } =\frac{3\times (-1)+2\times 4}{\sqrt{3^{2}+(-1)^{2} }\times \sqrt{2^{2}+4^{2} } }=\frac{5}{\sqrt{10}\times \sqrt{20} }=}

\latex{=\frac{5}{10\times \sqrt{2} }=\frac{1}{2\times \sqrt{2} }\approx 0.3536,}

which implies \latex{\varphi\approx 69.3°.}
The solution method of example \latex{ 4 } can be applied to determine the angle
of inclination of any two straight lines with the remark that if \latex{\cos \varphi\lt 0} (the angle included between the two direction vectors is an obtuse angle), then the angle of inclination of the two straight lines is \latex{180°-\varphi}.

Exercises

{{exercise_number}}. Calculate the coordinates of the intersection point of the straight lines \latex{ e } and \latex{ f }, if their equations are

\latex{e:x=2,f:y=2x-1;}

\latex{e:y=-4,f:x-y=5;}

\latex{e:y=x+1,f:y=3x+2;}

\latex{e:x-2=-1,f:3x+1=4.}

{{exercise_number}}. The straight lines of the sides of a triangle are : \latex{a:y=1,b:x+y=6,c:-5x+3y=15.} Calculate the coordinates of the vertices of the triangle.

{{exercise_number}}. Give the point of the straight line with the equation \latex{x+3y=7} that is equidistant from the points \latex{A(1;-9)} and \latex{B(-2;5).}

{{exercise_number}}. How far away is the point \latex{ P } from the straight line with the equation \latex{-x+2y=2,} if

\latex{P(0;0);}

\latex{P(0;-5);}

\latex{P(1;4);}

\latex{P(-3;-7);}

\latex{P(2;-6);}

\latex{P(4;3)?}

{{exercise_number}}. Calculate the lengths of the altitudes and the area of the triangle given in exercise \latex{ 2 }.

{{exercise_number}}. Calculate the distance of the parallel straight lines \latex{ e } and \latex{ f }, if their equations are

\latex{e:x=2,f:x=-4;}

\latex{e:y=-x,f:x+y=5;}

\latex{e:5x=-3y=-5,f:-5x+3y=15;}

\latex{e:3x+y=4,f:-9x-3y=-12.}

{{exercise_number}}. Calculate the area of the parallelogram three consecutive vertices of which labelled in positive orientation are \latex{A(-3;4),B(-2;-1),C(3;1).}

{{exercise_number}}. Give the angle of inclination of the pairs of straight lines given in exercise \latex{ 1 }.

{{exercise_number}}. Give the measures of the interior angles of the triangle given in exercise \latex{ 2 }.

{{exercise_number}}. The midpoints of the sides of a triangle are \latex{P(-2;3),Q(2;-1),R(4;6).} Calculate the lengths of the altitudes, the area and the measures of the interior angles of the triangle.

{{exercise_number}}. The vertices of a triangle are \latex{A(-4;6,)B(2;-4),C(6;5).}Calculate the coordinates of the intersection point of the median starting from the vertex \latex{ A } and the altitude starting from the vertex \latex{ B }, and also the area of the triangle.

{{exercise_number}}. The straight line e passes through the origin, \latex{\overrightarrow{v}(3;4)} is a direction vector of it; two points of the straight line \latex{ f } are \latex{A(2;6),B(15;-6).} Calculate the coordinates of the intersection point of the straight lines \latex{ e } and \latex{ f } and also their angle of inclination.

{{exercise_number}}. The equation of the straight line of one side of a regular triangle is \latex{x-2y=-6}, its centroid is \latex{S(5;-3).} Calculate the area of the triangle.

Mathematics 11.

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