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Ne - раздел Образование, Analytic Geometry In The Plane...
Analytic Geometry in the Plane
Consider the Cartesian rectangular coordinate system in the plane. Taking the projection of any point М1 on the x and y coordinate we obtain two… y a2 …Analytic geometry (all of its statements, theorems, and formulas) can be constructed on the basis elementary school mathematics. But we use tools of vector algebra in derivations and proofs.
The distance between two points. Let us find the distance between two points М1 and М2 in the plane.
M2(x2;,y2) d M1(x1;y1)Example. Find the distance between the two points А(2;3) and В(–4;11).
Using the above formula, we obtain
.
Division of an interval in a given ratio. Suppose given an interval М1М2. Let us find the coordinates a of point М on the interval for which .
y M2(x2,y2) M(x,y)Lines and Their Equations
yIntersection of lines. Suppose that lines and in the plane are given by their equations. How can we find the coordinates of their intersection point?
Parametric equations of lines. It is often difficult to compose an equation of a line in the plane in Cartesian coordinates, in the form .
yLet us express the sine and cosine as and from the first and the second equations. Squaring both parts and summing them, we obtain
, or 
, i.e.,
x2+y2=R2 ,
which coincides with the equation of the circle obtained above.
§ 2.7. Straight Lines in the Plane
The equation of a straight line with a slope. Given a straight line, we denote the angle between this line and the x–axis by j and the interval cut out by the line on the x-axis by b.
у М(х;у)
y-b
B j
b N
0 x
Definition. The slope tangent of the angle between a straight line and the x–axis is called the slope of the line and denoted by
k=tanj.
Suppose that k is the slope of a line and b is its y–intercept. Let us write an equation of this line.
Take a point M(x;y) and consider the triangle DBMN , where 
for any point М of the straight line under consideration. We obtain
, where 
and 
, whence 
.
Thus, the equation of a straight line with a slope has the form
.
2.7.2. The equation of a straight line with given slope passing through a given point.Suppose that a straight line passes through a point М0(x0,y0) and has slope k.
y
M(x;y)
φ
M0(x0;y0) N
φ
0 x
By analogy with the equation of a straight line with a slope consider the triangle М0MN; we have 
for any point М on the under
consideration or 
.
Thus, the required equation is
y – y0 = k(x–x0).
2.7.3. The equation of a straight line passing through two points. Suppose that a straight line passes through two points М1(х1;у1) and М2(х2;у2).
y
M(x;y)
M2(x2;y2)
M1(x1;y1)
0 x
Take a point M(x,y) on the line and consider the vectors
and 
.
These two vectors 
и 
lie on the same straight line and are collinear.
The collinearity condition is the proportionality of the perspective coordinates, i.e.,
(10)
This is the equation of a straight line passing through the two given points.
Example. Write an equation of the straight line passing through the points М1(2;–5) and М2(3;2) and find k and b.
Using formula (10), we obtain
Þ 7x–14=y+5.
Thus the equation of the straight line is
y=7x–19,
and the slope and the y–intercept are
k=7, b= –19.
2.7.4. The general equation of a straight line and its analysis. Definition. A first–order equation in variables x and y determines a straight line in the plane.
The general equation has the form
,
where А and В are called the coefficients of the variables.
1. If the free term is С=0, then the equation has the form
.
Since х=0 and y=0 satisfy this equation, it follows that the straight line passes through the origin.
2. If the coefficient of х is А=0, then the equation has the form
or 
, i.e., the line is parallel to the x–axis.
3. If the coefficient of y is B=0, then the equation has the form
,
and the line is parallel to the y–axis.
4. If А=С=0, then the line
В у=0 (or у=0)
coincides with the x–axis.
5. If В=С=0, then the line
А х=0 (or х=0)
coincides with the y–axis.
2.7. 5. The two-intercept equation of a straight line. Suppose that a straight line intersects the coordinate axes in points M1(a;0) and М2(0;b)
 |
у
М2(0;b)
b
M1(a;0)
0 a x
Knowing the two points М1 and М2 through which the line passes, we can write the equation of the line in form (10):
, or 
.
This is the two-intercept equation of the line.
The second derivation of two-intercept equation of straight line. Consider the general equation of a straight line:
Ax+By+C=0,
Where the coefficients A and B and the free term С are unknown.
Substituting the coordinates of the points M1(a;0 and М2(0;b) into this equation, we obtain
Аа+С=0,
Bb+C=0.
Therefore,
.
Substituting these coefficients into the general equation
and reducing by С, we obtain the two-intercept equation of the straight line.
Example. Reduce the equation 
, to the two-intercept form.
Take the variables to the left-hand side
.
We have
, where 
and 
2.7.6. The angle between two straight lines. Parallel and perpendicular lines.Suppose given two straight lines with slopes 
and 
.
у j=?
j1 j2
j1 j2
0 х
It is seen from the figure that the angle between the lines equals j=j2–j1. Using the formula for the tangent of the difference between
two angels, we obtain
.
Replacing the tangents by the slopes k1 and k2, we obtain the following formula for the tangent of the angle between two straight lines:
. (11)
Formula (11) gives conditions for two lines to be parallel and perpendicular.
(1) Suppose that the right lines are parallel, i.e., the angle between them is 
; substituting it into formula (11), we obtain
.
This fraction vanishes, if k2–k1=0.
Thus, two straight lines are parallel if and only if their slopes are equal:
k2=k1.
(2) Suppose that two straight lines are perpendicular; then the angle between them is 
. Substituting it in (11), we obtain
.
This fraction equals infinity when the denominator vanishes:
.
Consequently, the condition for two straight lines to be perpendicular is
.
Example. Write equations of the straight lines passing through the point М0(1;1) and parallel and perpendicular to the line 
.
Let us write the equation of the given line in the form y=kx+b:
, or 
; 
.
To compose an equation of a straight line, we use the formula
. (*)
Since the required line must be parallel to the required one, it follows that
.
Substituting this into equation (*), we obtain
, or 
.
The perpendicularity condition gives the slope:
.
The equation of the required line is
; 
.
2.7.7. The mutual arrangement of two straight lines. Given equations of two straight lines
and 
.
Determine conditions on the coefficient, for these right lines to intersect, be parallel, or coincide.
1. To determine the mutual arrangement of lines, we must analyze the system of equations
If the lines intersect, then this system has a unique solution, and its principal determinant is nonzero:
; 
;
. (12)
Thus, if the straight lines intersect, then the coefficients must not be proportional.
2. Suppose that the straight lines are parallel, i.e., they have no common points, and the system of equations has no solution; then the principal determinant vanishes, and the auxiliary determinants are nonzero:
; 
;
;
; 
; 
; 
;
. (12΄)
This is the parallelis condition.
3. When the straight lines coincide, i.e., have many common points, the system of equations has infinitely many solutions. In this case, the auxiliary and principal determinants are zero:
; 
;
. (12΄΄)
This is the condition for straight lines to coincide.
Example. 1. Concider the straight lines given by
According to formula (12), we have 
, which means that these lines intersect.
2. Consider the straight lines given by
According to formula (12΄), we have 
, and the lines are parallel.
3. Consider the straight lines given by
According to formula (12΄΄) we have 
, and the lines coincide.
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