What is the equation of a straight line. General equation of a straight line: description, examples, problem solving. Equation of a straight line from a point and a direction vector
The general equation of a second-order curve on a plane has the form:
Ax 2 + 2Bxy + Cy 2 + 2Dx + 2Ey + F = 0, (39)
Where A 2 + B 2 + C 2 0, (A, B, C, D, E, F) R. It defines all possible conic sections arbitrarily located on the plane.
From the coefficients of equation (39) we compose two determinants:
Called discriminant of the equation(39), and - discriminant of the leading terms of the equation. At 0, equation (39) determines: > 0 - ellipse;< 0 - гиперболу; = 0 - параболу. В случае = 0 кривые вырождаются в точку или прямые линии.
From the general equation (39) we can move to the canonical equation if we eliminate the linear and cross terms by moving to a new coordinate system that coincides with the symmetry axes of the figure. Let's replace in (39) x on x + a And y on y + b, Where a, b some constants. Let us write down the obtained coefficients for X And y and equate them to 0
(Aa + Bb + D)x = 0, (Cb + Ba + E)y = 0. (41)
As a result, equation (39) will take the form:
A(x) 2 + 2B(x)(y) + C(y) 2 + F = 0, (42)
where are the coefficients A, B, C haven't changed, but F= / . The solution to the system of equations (41) will determine the coordinates of the center of symmetry of the figure:
If B= 0, then a = -D/A, b = -E/C and it is convenient to eliminate linear terms in (39) by the method of reduction to a perfect square:
Ax 2 + 2Dx = A(x 2 + 2xD/A + (D/A) 2 - (D/A) 2) = A(x + D/A) 2 - D 2 /A.
In equation (42) we rotate the coordinates by angle a (38). Let us write down the resulting coefficient for the cross term xy and set it equal to 0
xy = 0. (44)
Condition (44) determines the required rotation angle of the coordinate axes until they coincide with the symmetry axes of the figure and takes the form:
Equation (42) takes the form:
A+X2+ C + Y 2 + F = 0 (46)
from which it is easy to go to the canonical equation of the curve:
Odds A + , C+ , under condition (45), can be represented as the roots of an auxiliary quadratic equation:
t 2 - (A + C)t + = 0. (48)
As a result, the position and direction of the axes of symmetry of the figure, its semi-axis, are determined:
and it can be constructed geometrically.
In the case = 0 we have a parabola. If its axis of symmetry is parallel to the axis Oh, then the equation reduces to:
if not, then look at:
where the expressions in brackets, equal to 0, define the lines of the new coordinate axes: , .
Solving common problems
Example 15. Give equation 2 x 2 + 3y 2 - 4x + 6y- 7 = 0 to canonical form and construct a curve.
Solution. B= 0, = -72 0, = 6 > 0 ellipse.
Let's perform a reduction to a perfect square:
2(x - 1) 2 + 3(y + 1) 2 - 12 = 0.
Coordinates of the center of symmetry (1; -1), linear transformation X = x - 1, Y = y+ 1 brings the equation to canonical form.
Example 16. Give equation 2 xy = a 2 to canonical form and construct a curve.
Solution. B = 1, = a 2 0, = -1 < 0 гипербола .
The center of the coordinate system is at the center of symmetry of the curve, because there are no linear terms in the equation. Let's rotate the axes by an angle a. According to formula (45) we have tan2a = B/(A - C) = , i.e. a = 45°. Coefficients of the canonical equation (46) A + , C+ are determined by equation (48): t 2 = 1 or t 1,2 = 1 A + = 1, C+ = -1, i.e.
X 2 - Y 2 = a 2 or . So equation 2 xy = A 2 describes a hyperbola with the center of symmetry at (0; 0). The axes of symmetry are located along the bisectors of the coordinate angles, the coordinate axes serve as asymptotes, the semi-axes of the hyperbola are equal A.y - 9 =0;
9x 2 + y 2 - 18x + 2y + 1 = 0;
2x 2 + 4X + y - 2 = 0;
3x 2 - 6X - y + 2 = 0;
-x 2 + 4y 2 - 8x - 9y + 16 = 0;
4x 2 + 8X - y - 5 = 0;
9x 2 - y 2 + 18x + 2y - 1 = 0;
9x 2 - 4y 2 + 36x + 16y - 16 = 0.
As shown above, equations of the same line can be written in at least three forms: general equations of the line, parametric equations of the line, and canonical equations of the line. Let us consider the question of the transition from straight line equations of one type to straight line equations in another form.
First, we note that if the equations of a line are given in parametric form, then the point through which the line passes and the direction vector of the line are thereby given. Therefore, it is not difficult to write down the equations of a straight line in canonical form.
Example.
The equations of the line are given in parametric form
Solution.
A straight line passes through a point 
and has a direction vector 
. Consequently, the canonical equations of the line have the form
.
The problem of transition from the canonical equations of the line to the parametric equations of the line is solved in a similar way.
The transition from the canonical equations of the line to the general equations of the line is discussed below using an example.
Example.
The canonical equations of the line are given
.
Write down the general equations of the line.
Solution.
Let us write the canonical equations of the line in the form of a system of two equations
.
Getting rid of the denominators by multiplying both sides of the first equation by 6, and the second equation by 4, we get the system
.
.
The resulting system of equations is the general equations of the straight line.
Let us consider the transition from general equations of the line to parametric and canonical equations of the line. To write the canonical or parametric equations of a line, you need to know the point through which the line passes and the direction vector of the line. If we determine the coordinates of two points 
And 
, lying on a straight line, then the vector m can be taken as the direction vector 
. The coordinates of two points lying on a line can be obtained as solutions to a system of equations that determine the general equations of the line. You can take any of the points as the point through which the line passes 
And 
. Let us illustrate the above with an example.
Example.
The general equations of the straight line are given
.
Solution.
Let's find the coordinates of two points lying on a straight line as solutions to this system of equations. Believing 
, we obtain a system of equations
.
Solving this system, we find 
. Therefore, the point 
lies on a straight line. Believing 
, we obtain a system of equations
,
solving which we find 
. Therefore, the straight line passes through the point 
. Then we can take the vector as the direction vector
.
So the line passes through the point 
and has a direction vector 
. Consequently, the parametric equations of the line have the form
.
Then the canonical equations of the line will be written in the form
.
Another way to find the direction vector of a straight line using the general equations of a straight line is based on the fact that in this case the equations of the planes are given, and hence the normals to these planes.
Let the general equations of the line have the form
And 
- normals to the first and second planes, respectively. Then the vector 
can be taken as a directing vector. In fact, the straight line, being the line of intersection of these planes, is simultaneously perpendicular to the vectors 
And 
. Therefore, it is collinear to the vector 
and this means that this vector can be taken as the directing vector of the straight line. Let's look at an example.
Example.
The general equations of the straight line are given
.
Write down the parametric and canonical equations of the line.
Solution.
The straight line is the line of intersection of planes with normals 
And 
. We take the direct vector as the direction vector
Let's find a point lying on a line. Let's find a point lying on a line. Let 
. Then we get the system
.
Solving the system, we find 
.Hence, period 
lies on a straight line. Then the parametric equations of the line can be written in the form
.
The canonical equations of the line have the form
.
Finally, we can move to canonical equations by eliminating one of the variables in one of the equations, and then another variable. Let's look at this method with an example.
Example.
The general equations of the straight line are given
.
Write down the canonical equations of the line.
Solution.
Let's exclude the variable y from the second equation by adding to it the first one, multiplied by four. We get
.
.
Now let's exclude the variable from the second equation 
, adding to it the first equation multiplied by two. We get
.
.
From here we obtain the canonical equation of the line
.
.
.
Second order curve— geometric location of points on the plane, rectangular coordinates
which satisfy an equation of the form:
in which at least one of the coefficients a 11, a 12, a 22 not equal to zero.
Invariants of second order curves.
The shape of the curve depends on 4 invariants given below:
Invariants with respect to rotation and shift of the coordinate system:
Invariant with respect to rotation of the coordinate system ( semi-invariant):
To study second-order curves, consider the product A*S.
General second order curve equation looks like that:
Ax 2 +2Bxy+Cy 2 +2Dx+2Ey+F=0
If A*C > 0 elliptical type. Any elliptical
equation is the equation of either an ordinary ellipse, or a degenerate ellipse (point), or an imaginary one
ellipse (in this case the equation does not define a single geometric image on the plane);
If A*C< 0 , then the equation takes the form of equation hyperbolic type. Any hyperbolic
the equation expresses either a simple hyperbola or a degenerate hyperbola (two intersecting lines);
If A*C = 0, then the second-order line will not be central. Equations of this type are called
equations parabolic type and express on the plane either a simple parabola, or 2 parallel
(either coinciding) straight lines, or do not express a single geometric image on the plane;
If A*C ≠ 0, the second order curve will be
This article continues the topic of the equation of a line on a plane: we will consider this type of equation as the general equation of a line. Let us define the theorem and give its proof; Let's figure out what an incomplete general equation of a line is and how to make transitions from a general equation to other types of equations of a line. We will reinforce the entire theory with illustrations and solutions to practical problems.
Let a rectangular coordinate system O x y be specified on the plane.
Theorem 1
Any equation of the first degree, having the form A x + B y + C = 0, where A, B, C are some real numbers (A and B are not equal to zero at the same time), defines a straight line in a rectangular coordinate system on a plane. In turn, any straight line in a rectangular coordinate system on a plane is determined by an equation that has the form A x + B y + C = 0 for a certain set of values A, B, C.
Proof
This theorem consists of two points; we will prove each of them.
- Let us prove that the equation A x + B y + C = 0 defines a straight line on the plane.
Let there be some point M 0 (x 0 , y 0) whose coordinates correspond to the equation A x + B y + C = 0. Thus: A x 0 + B y 0 + C = 0. Subtract from the left and right sides of the equations A x + B y + C = 0 the left and right sides of the equation A x 0 + B y 0 + C = 0, we obtain a new equation that looks like A (x - x 0) + B (y - y 0) = 0 . It is equivalent to A x + B y + C = 0.
The resulting equation A (x - x 0) + B (y - y 0) = 0 is a necessary and sufficient condition for the perpendicularity of the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0 ) . Thus, the set of points M (x, y) defines a straight line in a rectangular coordinate system perpendicular to the direction of the vector n → = (A, B). We can assume that this is not so, but then the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) would not be perpendicular, and the equality A (x - x 0 ) + B (y - y 0) = 0 would not be true.
Consequently, the equation A (x - x 0) + B (y - y 0) = 0 defines a certain line in a rectangular coordinate system on the plane, and therefore the equivalent equation A x + B y + C = 0 defines the same line. This is how we proved the first part of the theorem.
- Let us present a proof that any straight line in a rectangular coordinate system on a plane can be specified by an equation of the first degree A x + B y + C = 0.
Let us define a straight line a in a rectangular coordinate system on a plane; the point M 0 (x 0 , y 0) through which this line passes, as well as the normal vector of this line n → = (A, B) .
Let there also be some point M (x, y) - a floating point on a line. In this case, the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) are perpendicular to each other, and their scalar product is zero:
n → , M 0 M → = A (x - x 0) + B (y - y 0) = 0
Let's rewrite the equation A x + B y - A x 0 - B y 0 = 0, define C: C = - A x 0 - B y 0 and as a final result we get the equation A x + B y + C = 0.
So, we have proved the second part of the theorem, and we have proved the entire theorem as a whole.
Definition 1
An equation of the form A x + B y + C = 0 - This general equation of a line on a plane in a rectangular coordinate systemOxy.
Based on the proven theorem, we can conclude that a straight line and its general equation defined on a plane in a fixed rectangular coordinate system are inextricably linked. In other words, the original line corresponds to its general equation; the general equation of a line corresponds to a given line.
From the proof of the theorem it also follows that the coefficients A and B for the variables x and y are the coordinates of the normal vector of the line, which is given by the general equation of the line A x + B y + C = 0.
Let's consider a specific example of a general equation of a straight line.
Let the equation 2 x + 3 y - 2 = 0 be given, which corresponds to a straight line in a given rectangular coordinate system. The normal vector of this line is the vector n → = (2 , 3) . Let's draw the given straight line in the drawing.
We can also state the following: the straight line that we see in the drawing is determined by the general equation 2 x + 3 y - 2 = 0, since the coordinates of all points on a given straight line correspond to this equation.
We can obtain the equation λ · A x + λ · B y + λ · C = 0 by multiplying both sides of the general equation of the line by a number λ not equal to zero. The resulting equation is equivalent to the original general equation, therefore, it will describe the same straight line on the plane.
Definition 2Complete general equation of a line– such a general equation of the straight line A x + B y + C = 0, in which the numbers A, B, C are different from zero. Otherwise the equation is incomplete.
Let us analyze all variations of the incomplete general equation of a line.
- When A = 0, B ≠ 0, C ≠ 0, the general equation takes the form B y + C = 0. Such an incomplete general equation defines in a rectangular coordinate system O x y a straight line that is parallel to the O x axis, since for any real value of x the variable y will take the value - C B . In other words, the general equation of the straight line A x + B y + C = 0, when A = 0, B ≠ 0, specifies the locus of points (x, y), whose coordinates are equal to the same number - C B .
- If A = 0, B ≠ 0, C = 0, the general equation takes the form y = 0. This incomplete equation defines the x-axis O x .
- When A ≠ 0, B = 0, C ≠ 0, we obtain an incomplete general equation A x + C = 0, defining a straight line parallel to the ordinate.
- Let A ≠ 0, B = 0, C = 0, then the incomplete general equation will take the form x = 0, and this is the equation of the coordinate line O y.
- Finally, for A ≠ 0, B ≠ 0, C = 0, the incomplete general equation takes the form A x + B y = 0. And this equation describes a straight line that passes through the origin. In fact, the pair of numbers (0, 0) corresponds to the equality A x + B y = 0, since A · 0 + B · 0 = 0.
Let us graphically illustrate all of the above types of incomplete general equation of a straight line.
Example 1
It is known that the given straight line is parallel to the ordinate axis and passes through the point 2 7, - 11. It is necessary to write down the general equation of the given line.
Solution
A straight line parallel to the ordinate axis is given by an equation of the form A x + C = 0, in which A ≠ 0. The condition also specifies the coordinates of the point through which the line passes, and the coordinates of this point meet the conditions of the incomplete general equation A x + C = 0, i.e. the equality is true:
A 2 7 + C = 0
From it it is possible to determine C if we give A some non-zero value, for example, A = 7. In this case, we get: 7 · 2 7 + C = 0 ⇔ C = - 2. We know both coefficients A and C, substitute them into the equation A x + C = 0 and get the required straight line equation: 7 x - 2 = 0
Answer: 7 x - 2 = 0
Example 2
The drawing shows a straight line; you need to write down its equation.
Solution
The given drawing allows us to easily take the initial data to solve the problem. We see in the drawing that the given straight line is parallel to the O x axis and passes through the point (0, 3).
The straight line, which is parallel to the abscissa, is determined by the incomplete general equation B y + C = 0. Let's find the values of B and C. The coordinates of the point (0, 3), since the given line passes through it, will satisfy the equation of the line B y + C = 0, then the equality is valid: B · 3 + C = 0. Let's set B to some value other than zero. Let's say B = 1, in which case from the equality B · 3 + C = 0 we can find C: C = - 3. Using the known values of B and C, we obtain the required equation of the straight line: y - 3 = 0.
Answer: y - 3 = 0 .
General equation of a line passing through a given point in a plane
Let the given line pass through the point M 0 (x 0 , y 0), then its coordinates correspond to the general equation of the line, i.e. the equality is true: A x 0 + B y 0 + C = 0. Let us subtract the left and right sides of this equation from the left and right sides of the general complete equation of the line. We get: A (x - x 0) + B (y - y 0) + C = 0, this equation is equivalent to the original general one, passes through the point M 0 (x 0, y 0) and has a normal vector n → = (A, B) .
The result that we obtained makes it possible to write down the general equation of a line with known coordinates of the normal vector of the line and the coordinates of a certain point of this line.
Example 3
Given a point M 0 (- 3, 4) through which a line passes, and the normal vector of this line n → = (1 , - 2) . It is necessary to write down the equation of the given line.
Solution
The initial conditions allow us to obtain the necessary data to compose the equation: A = 1, B = - 2, x 0 = - 3, y 0 = 4. Then:
A (x - x 0) + B (y - y 0) = 0 ⇔ 1 (x - (- 3)) - 2 y (y - 4) = 0 ⇔ ⇔ x - 2 y + 22 = 0
The problem could have been solved differently. The general equation of a line is A x + B y + C = 0. The given normal vector allows us to obtain the values of coefficients A and B, then:
A x + B y + C = 0 ⇔ 1 x - 2 y + C = 0 ⇔ x - 2 y + C = 0
Now let’s find the value of C using the point M 0 (- 3, 4) specified by the problem condition, through which the straight line passes. The coordinates of this point correspond to the equation x - 2 · y + C = 0, i.e. - 3 - 2 4 + C = 0. Hence C = 11. The required straight line equation takes the form: x - 2 · y + 11 = 0.
Answer: x - 2 y + 11 = 0 .
Example 4
Given a line 2 3 x - y - 1 2 = 0 and a point M 0 lying on this line. Only the abscissa of this point is known, and it is equal to - 3. It is necessary to determine the ordinate of a given point.
Solution
Let us designate the coordinates of point M 0 as x 0 and y 0 . The source data indicates that x 0 = - 3. Since the point belongs to a given line, then its coordinates correspond to the general equation of this line. Then the equality will be true:
2 3 x 0 - y 0 - 1 2 = 0
Define y 0: 2 3 · (- 3) - y 0 - 1 2 = 0 ⇔ - 5 2 - y 0 = 0 ⇔ y 0 = - 5 2
Answer: - 5 2
Transition from the general equation of a line to other types of equations of a line and vice versa
As we know, there are several types of equations for the same straight line on a plane. The choice of the type of equation depends on the conditions of the problem; it is possible to choose the one that is more convenient for solving it. The skill of converting an equation of one type into an equation of another type is very useful here.
First, let's consider the transition from the general equation of the form A x + B y + C = 0 to the canonical equation x - x 1 a x = y - y 1 a y.
If A ≠ 0, then we move the term B y to the right side of the general equation. On the left side we take A out of brackets. As a result, we get: A x + C A = - B y.
This equality can be written as a proportion: x + C A - B = y A.
If B ≠ 0, we leave only the term A x on the left side of the general equation, transfer the others to the right side, we get: A x = - B y - C. We take – B out of brackets, then: A x = - B y + C B .
Let's rewrite the equality in the form of a proportion: x - B = y + C B A.
Of course, there is no need to memorize the resulting formulas. It is enough to know the algorithm of actions when moving from a general equation to a canonical one.
Example 5
The general equation of the line 3 y - 4 = 0 is given. It is necessary to transform it into a canonical equation.
Solution
Let's write the original equation as 3 y - 4 = 0. Next, we proceed according to the algorithm: the term 0 x remains on the left side; and on the right side we put - 3 out of brackets; we get: 0 x = - 3 y - 4 3 .
Let's write the resulting equality as a proportion: x - 3 = y - 4 3 0 . Thus, we have obtained an equation of canonical form.
Answer: x - 3 = y - 4 3 0.
To transform the general equation of a line into parametric ones, first a transition is made to the canonical form, and then a transition from the canonical equation of a line to parametric equations.
Example 6
The straight line is given by the equation 2 x - 5 y - 1 = 0. Write down the parametric equations for this line.
Solution
Let us make the transition from the general equation to the canonical one:
2 x - 5 y - 1 = 0 ⇔ 2 x = 5 y + 1 ⇔ 2 x = 5 y + 1 5 ⇔ x 5 = y + 1 5 2
Now we take both sides of the resulting canonical equation equal to λ, then:
x 5 = λ y + 1 5 2 = λ ⇔ x = 5 λ y = - 1 5 + 2 λ , λ ∈ R
Answer:x = 5 λ y = - 1 5 + 2 λ , λ ∈ R
The general equation can be converted into an equation of a straight line with slope y = k · x + b, but only when B ≠ 0. For the transition, we leave the term B y on the left side, the rest are transferred to the right. We get: B y = - A x - C . Let's divide both sides of the resulting equality by B, different from zero: y = - A B x - C B.
Example 7
The general equation of the line is given: 2 x + 7 y = 0. You need to convert that equation into a slope equation.
Solution
Let's perform the necessary actions according to the algorithm:
2 x + 7 y = 0 ⇔ 7 y - 2 x ⇔ y = - 2 7 x
Answer: y = - 2 7 x .
From the general equation of a line, it is enough to simply obtain an equation in segments of the form x a + y b = 1. To make such a transition, we move the number C to the right side of the equality, divide both sides of the resulting equality by – C and, finally, transfer the coefficients for the variables x and y to the denominators:
A x + B y + C = 0 ⇔ A x + B y = - C ⇔ ⇔ A - C x + B - C y = 1 ⇔ x - C A + y - C B = 1
Example 8
It is necessary to transform the general equation of the line x - 7 y + 1 2 = 0 into the equation of the line in segments.
Solution
Let's move 1 2 to the right side: x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2 .
Let's divide both sides of the equality by -1/2: x - 7 y = - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y = 1 .
Answer: x - 1 2 + y 1 14 = 1 .
In general, the reverse transition is also easy: from other types of equations to the general one.
The equation of a straight line in segments and an equation with an angular coefficient can be easily converted into a general one by simply collecting all the terms on the left side of the equality:
x a + y b ⇔ 1 a x + 1 b y - 1 = 0 ⇔ A x + B y + C = 0 y = k x + b ⇔ y - k x - b = 0 ⇔ A x + B y + C = 0
The canonical equation is converted to a general one according to the following scheme:
x - x 1 a x = y - y 1 a y ⇔ a y · (x - x 1) = a x (y - y 1) ⇔ ⇔ a y x - a x y - a y x 1 + a x y 1 = 0 ⇔ A x + B y + C = 0
To move from parametric ones, first move to the canonical one, and then to the general one:
x = x 1 + a x · λ y = y 1 + a y · λ ⇔ x - x 1 a x = y - y 1 a y ⇔ A x + B y + C = 0
Example 9
The parametric equations of the line x = - 1 + 2 · λ y = 4 are given. It is necessary to write down the general equation of this line.
Solution
Let us make the transition from parametric equations to canonical ones:
x = - 1 + 2 · λ y = 4 ⇔ x = - 1 + 2 · λ y = 4 + 0 · λ ⇔ λ = x + 1 2 λ = y - 4 0 ⇔ x + 1 2 = y - 4 0
Let's move from the canonical to the general:
x + 1 2 = y - 4 0 ⇔ 0 · (x + 1) = 2 (y - 4) ⇔ y - 4 = 0
Answer: y - 4 = 0
Example 10
The equation of a straight line in the segments x 3 + y 1 2 = 1 is given. It is necessary to transition to the general form of the equation.
Solution:
We simply rewrite the equation in the required form:
x 3 + y 1 2 = 1 ⇔ 1 3 x + 2 y - 1 = 0
Answer: 1 3 x + 2 y - 1 = 0 .
Drawing up a general equation of a line
We said above that the general equation can be written with known coordinates of the normal vector and the coordinates of the point through which the line passes. Such a straight line is defined by the equation A (x - x 0) + B (y - y 0) = 0. There we also analyzed the corresponding example.
Now let's look at more complex examples, in which first we need to determine the coordinates of the normal vector.
Example 11
Given a line parallel to the line 2 x - 3 y + 3 3 = 0. The point M 0 (4, 1) through which the given line passes is also known. It is necessary to write down the equation of the given line.
Solution
The initial conditions tell us that the lines are parallel, then, as the normal vector of the line, the equation of which needs to be written, we take the direction vector of the line n → = (2, - 3): 2 x - 3 y + 3 3 = 0. Now we know all the necessary data to create the general equation of the line:
A (x - x 0) + B (y - y 0) = 0 ⇔ 2 (x - 4) - 3 (y - 1) = 0 ⇔ 2 x - 3 y - 5 = 0
Answer: 2 x - 3 y - 5 = 0 .
Example 12
The given line passes through the origin perpendicular to the line x - 2 3 = y + 4 5. It is necessary to create a general equation for a given line.
Solution
The normal vector of a given line will be the direction vector of the line x - 2 3 = y + 4 5.
Then n → = (3, 5) . The straight line passes through the origin, i.e. through point O (0, 0). Let's create a general equation for a given straight line:
A (x - x 0) + B (y - y 0) = 0 ⇔ 3 (x - 0) + 5 (y - 0) = 0 ⇔ 3 x + 5 y = 0
Answer: 3 x + 5 y = 0 .
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