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The equation of a straight line I

We saw it earlier that a straight line is unambiguously defined by the
following in the coordinate system:

one point and a direction vector;
one point and a normal vector;
one point and its angle of inclination;
one point and its gradient (if it exists);
two points.

Our aim now is to find a relation between the coordinates of the points of the straight line, which holds for all the points of the straight line but does not hold for the points not lying on the straight line.

Example 1

A point of the straight line e is \latex{P_{0}(2;1)}, a normal vector is \latex{\overrightarrow{n}(5;2).} What is the relation between the coordinates of the point \latex{P(x;y)} if \latex{ P } lies on the straight line?

\latex{\overrightarrow{P_{0}P }}

\latex{\overrightarrow{r}}

\latex{\overrightarrow{r_{0} }}

\latex{\overrightarrow{P_{0}P }=\overrightarrow{r}-\overrightarrow{r_{0} }}

\latex{P_{0}(2;1)}

\latex{\overrightarrow{n}(5;2)}

\latex{P(x;y)}

\latex{ 0 }

\latex{ 1 }

\latex{ e }

\latex{ y }

\latex{ x }

Figure 27

Solution

The definition of the normal vector implies that the point \latex{ P } lies on the straight line \latex{ e } if and only if the vectors \latex{\overrightarrow{P_{0}P }} and \latex{\overrightarrow{n}} are perpendicular to each other (Figure 27), and it is satisfied exactly when their dot product is \latex{ 0 }. If \latex{\overrightarrow{r_{0} }} is the position vector of \latex{P_{0}} and \latex{\overrightarrow{r}} is the position vector of \latex{ P }, then \latex{\overrightarrow{P_{0}P }=\overrightarrow{r}-\overrightarrow{r_{0} }.}

By expressing the dot product of the two vectors with coordinates

\latex{\overrightarrow{n}\times \overrightarrow{P_{0}P }=\overrightarrow{n}\times (\overrightarrow{r}-\overrightarrow{r_{0} } )=5\times (x-2)+2\times (y-1)=0} which implies
\latex{5x+2y=12.}

The coordinates of all the points of the straight line e satisfy this equation, and the coordinates of no other points satisfy it. So the relation in question is obtained.

DEFINITION: The equation of a figure located in the planar Cartesian coordinate system is an equation in two variables that is satisfied by the coordinates of the points \latex{P(x;y)} of the figure, but it is not satisfied by the coordinates of any other point.

The method of the solution of the first example can be generalised:
The point \latex{P_{0}(x_{0};y_{0} )} and the normal vector \latex{\overrightarrow{n}(A;B)} of the straight line e are given. The point \latex{P(x;y)} lies on the straight line e if and only if

\latex{\overrightarrow{n}\times \overrightarrow{P_{0}P }=0.}

Let the position vector of the fixed point \latex{P_{0}} be denoted by \latex{\overrightarrow{r_{0} }}, and let the position vector of the arbitrary point \latex{ P } (variable point) be denoted by \latex{\overrightarrow{r}} (Figure 28).

With this notation

\latex{\overrightarrow{n}\times (\overrightarrow{r}-\overrightarrow{r_{0} } ) =0.}

This is the vector equation of the straight line.
By expressing with coordinates

\latex{\overrightarrow{n}\times (\overrightarrow{r}-\overrightarrow{r_{0} } )=A\times (x-x_{0} )+B\times (y-y_{0} )=0,}

which implies

\latex{Ax+By=Ax_{0}+By_{0}.}

The relation obtained is the equation of the straight line given by the normal vector \latex{\overrightarrow{n}(A;B)} and the point \latex{P_{0}(x_{0};y_{0} ),} or in other words the general form of the equation of the straight line.

Example 2

A point of the straight line g is \latex{P_{0}(3;-4),} a normal vector is \latex{\overrightarrow{n}(-1;1)}. Let us give the equation of the straight line, and let us determine the intersection points of \latex{ g } and the coordinate axes.

Solution

As \latex{A=-1,B=1,x_{0}=3,y_{0}=-4}, the equation of the straight line \latex{ g } is:

\latex{-x+y=(-1)\times 3+1\times (-4)=-7,}

i.e.

\latex{-x+y=-7.}

The straight line g intersects the \latex{ x }-axis at the point \latex{ P } the second coordinate of which is \latex{ 0 }. By substituting it into the equation \latex{x=7,} is obtained, i.e. \latex{ g } intersects the \latex{ x }-axis at the point \latex{P(7;0).}

The first coordinate of the point \latex{ Q } of the straight line \latex{ g } also lying on the \latex{ y }-axis is \latex{ 0 }; its second coordinate is obtained from the equation: \latex{y=-7.} So \latex{ g } intersects the \latex{ y }-axis at the point \latex{Q(0;-7)} (Figure 29).

Example 3

The equation of the straight line \latex{ f } is \latex{6y-3x+12=0.} Let us give a normal vector, a direction vector, the gradient and the first coordinate of the point \latex{ P } of the straight line the second coordinate of which is \latex{ 4 }.

Solution

Let us first rearrange the equation to a form in which the coordinates of a normal vector can easily be seen.

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

By comparing it to the general form it can be seen that \latex{\overrightarrow{n_{1} }(-3;6)} is a normal vector of \latex{ f }. Based on a remark we made earlier \latex{\overrightarrow{n_{2} }(-1;2)} is also a normal vector of \latex{ f }, which can also be obtained by dividing both sides of the equation by \latex{ 3 }:

\latex{-x+2y=-4.}

Based on the relation between the direction characteristics of the straight line \latex{\overrightarrow{v}(2;1)} is a direction vector, and \latex{m=\frac{1}{2}} is the gradient of \latex{ f }.
The first coordinate of P is the solution of the equation \latex{-x+2\times 4=-4,} i.e. \latex{x=12.}

Example 4

Let us give the equation of the perpendicular bisector of the line segment \latex{ AB }, if \latex{A(-2;-1)} and \latex{B(6;5)}.

Solution

Let \latex{f_{AB}} denote the perpendicular bisector in question. \latex{f_{AB}} passes through the midpoint \latex{ F } of the line segment \latex{ AB }, and it is perpendicular to \latex{ AB } (Figure 30), thus the point \latex{ F } and \latex{\overrightarrow{AB}} as a normal vector unambiguously define \latex{f_{AB}}.
Based on the relation relating to the coordinates of the midpoint of a line segment both coordinates of \latex{ F } are \latex{ 2 }, and

\latex{\overrightarrow{AB}=(6-(-2))i+(5-(-1))j=8i+6j.}

For the sake of simpler calculation it is worth taking half of \latex{\overrightarrow{AB}}, i.e. the normal vector \latex{\overrightarrow{n}(4;3)} to give the equation. The equation of the straight line \latex{f_{AB}} passing through the point \latex{F(2;2)} and with the normal vector \latex{\overrightarrow{n}(4;3)}, by substituting into the general form, is

\latex{4x+3y=14.}

Exercises

{{exercise_number}}. Give the equation of the straight line passing through the origin and with the normal vector \latex{\overrightarrow{n}} in the cases below.

\latex{\overrightarrow{n}(1;1)}

\latex{\overrightarrow{n}(2;3)}

\latex{\overrightarrow{n}(-1;4)}

\latex{\overrightarrow{n}(5;-3)}

\latex{\overrightarrow{n}(-4;-8)}

{{exercise_number}}. Give the equation of the straight line passing through the point \latex{P_{0}} and with the normal vector \latex{\overrightarrow{n}}, if

\latex{P_{0}(0;1),\overrightarrow{n}(-1;1)}

\latex{P_{0}(-2;0),\overrightarrow{n}(3;1)}

\latex{P_{0}(3;-4),\overrightarrow{n}(-2;8)}

{{exercise_number}}. Give a normal vector, a direction vector and the gradient (if it exists) of the straight line the equation of which is

\latex{y=4;}

\latex{x+y=-2;}

\latex{-x+3y=11;}

\latex{y+7=4x-2;}

\latex{x=2.}

{{exercise_number}}. The point \latex{ P } lies on the straight line defined by the point \latex{P_{0}(1;-4)} and the normal vector \latex{\overrightarrow{n}(5;2).} Calculate the second coordinate of \latex{ P }, if its first coordinate is.

\latex{0}

\latex{1}

\latex{-2}

\latex{5}

\latex{-\frac{2}{3}}

\latex{\sqrt{5}}

{{exercise_number}}. Give the equation of the perpendicular bisector of the line segment \latex{ AB }, if

\latex{A(0;1),B(1;0)};

\latex{A(2;4),B(1;-3)};

\latex{A(-4;5),B(2;0)};

\latex{A(-7;-8),B(3;-5)}.

Mathematics 11.

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