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Line Pencils

Definition. A line pencil is the set of straight line passing through a given point.

Arbitrarily varying the coefficient k in the equation of a line passing through a point 0(0,0), we obtain equations of all straight lines passing through the point with coordinates 0, 0.

All these straight lines constitute the line pencil centered at 0.

A line pencil can also be specified by two equations of lines from this pencil (which determine of these lines) the common point 0(0, 0) by the coordinates of its center.

Suppose given intersecting straight lines with general equations

,

.

Consider the equations

, (13)

where q1 and q2 undetermined multipliers, not both zero.

Let us prove that, for any q1 and q2, the straight line (13) passes through the point 0 (0; 0), which is the intersection point of the two given lines.

The coordinates of the point 0 satisfy both equations

,

Consequently, the coordinates of 0satisfy equation (13), i.e.,

.

Thus, the point 0 belongs to the straight line (13) for any q1 and q2; but for particular values of q1 and q2, we obtain some straight line from the pencil.

In practice, it is more convenient to write an equation of a pencil with one parameter l:

.

Example. Draw a straight line through the intersection point of the lines +3=0 and 2+5=0 so that it passes through the point (3;5).

Let us write the equation of the pencil of lines according to the formula. To find the required line, we must find the value of the parameter l:

We substitute the coordinates of the point (3;5) into the equation of the pencil:

,

.

Substituting l in to the equation of the pencil, we obtain the required equation:

;

.

The Normal Equation of a Straight Line

 

The position of a straight line in the plane is completely determined by the distance from the line to the origin of coordinates, i.e., the length of the perpendicular from the origin to the fight line, and the unit normal vector

y

 

 

α M

0 x

 

 

Let us construct an equation of a straight line from the length of the perpendicular and its angle a with the x-axis.

To this end we take an arbitrary point on the line and find the projection of the vector on the unit normal vector .

The point belongs to the given line if and only if

. (*)

In terms of inner product of vectors, we have

.

Considering this product, we obtain

. (14)

Equation (14) is the normal equation of the line in vector form. To pass to coordinates, note that the projections of the unit normal vector are

,

where a is the angle between this vector and the axis X. Let the projections of the radius-vectors on the axes and are and , respectively. The inner product formula (14) implies

. (15)

This equation is called the normal equation of the straight line. It is clear from equation (15) that the normality conditions are

(1) (the sum of squared coefficients of and equals 1);



(2) <0 (the free element is negative).

Example. Which of the following equations are normal?

(a) , , this is not a normal equation;

(b) , and 4<0, this is a normal equation;

(c) , but 2>0, this is not a normal equation.

 

A normalizing factor. Suppose given a general equation of a straight line given:

.

Definition. The normalizing factor is the number such that the equation multiplied by this number is normal.

Let us multiply the general equation of a straight line by number m:

.

By definition, if this equation is normal, then the two normality conditions hold:

; .

Removing parentheses, we obtain

;

extracting the root, we see that the normalizing factor is

;

the sign of the fraction is opposite to that of the free element of the general equation.

Example. Reduce the equation 86+5=0 to the normal form.

We find the normalizing factor

and multiply the equation by this factor term by term:

or .

 


Date: 2015-01-02; view: 788


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The distance between two points. Let us find the distance between two points 1 and 2 in the plane. | Analytic geometry in space. Vectors. Simple operations with vectors. The scalar, vector and mixed product of vectors
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