**Line angle coefficient** - coefficient characterizing the degree of slope of the line. Coefficient * k* in the equation

*line on the coordinate plane, numerically equal to the tangent of the angle (making up the smallest rotation from the Ox axis to the Oy axis) between the positive direction of the abscissa axis and this straight line*

**y = kx + b**### Problem: Find the slope of the line given by equation 36x - 18y = 108

Solution: Transform the original equation.

Answer: The desired slope of this line is 2.

If during the transformations of the equation we get an expression of the type x = const and cannot as a result represent y as a function of x, then we are dealing with a line parallel to the X axis. The angular coefficient of such a line is infinity.

For lines that are expressed by an equation of type y = const, the slope is zero. This is typical for straight lines parallel to the abscissa axis. For example:

## Geometric meaning

For a better understanding, refer to the picture:

In the figure, we see a graph of a function of type y = kx. For simplicity, we take the coefficient c = 0. In the triangle OAV, the ratio of the side of the VA to the AO will be equal to the angular coefficient k. At the same time, the VA / AO ratio is the tangent of an acute angle α in a right-angled triangle of OAV. It turns out that the angular coefficient of the line is equal to the tangent of the angle that this line makes up with the abscissa axis of the coordinate grid.

Solving the problem of how to find the angular coefficient of a straight line, we find the tangent of the angle between it and the X axis of the coordinate grid. Boundary cases when the straight line in question is parallel to the coordinate axes confirm the above. Indeed, for the line described by the equation y = const, the angle between it and the abscissa axis is equal to zero. The tangent of the zero angle is also equal to zero and the angular coefficient is also equal to zero.

For straight lines perpendicular to the abscissa axis and described by the equation x = const, the angle between them and the X axis is 90 degrees. The tangent of a right angle is equal to infinity, as well as the angular coefficient of such lines is equal to infinity, which confirms the above.

## Tangent angle coefficient

A common, often encountered in practice, task is also to find the angular coefficient of the tangent to the graph of the function at some point. The tangent is a line, therefore the concept of an angular coefficient is also applicable to it.

To understand how to find the angular coefficient of a tangent, we will need to recall the concept of a derivative. The derivative of any function at some point is a constant numerically equal to the tangent of the angle that is formed between the tangent at the specified point to the graph of this function and the abscissa axis. It turns out that to determine the angular coefficient of the tangent at the point x_{0}, we need to calculate the value of the derivative of the original function at this point k = f '(x_{0}) Consider an example:

### Angle Ratio Calculator

Online calculator for finding the angular coefficient of a straight line. Easily find your slope online

- A straight line is increasing if it goes up from left to right. The angle coefficient is positive,
**k****>0.** - A straight line decreases if it goes down from left to right. The angular coefficient is negative,
**k k = (Y2 - Y1) / (X2 - X1)**

(or)**k = (Y1 - Y2) / (X1 - X2)**

- k = Angular coefficient of the line,
- X1, X2 = X coordinates,
- Y1, Y2 = Y coordinates.

### An example of finding the angular coefficient:

Suppose a line passes through two points: *P* = (1, 2) and *Q* = (13, 8). Dividing the difference *y*-coordinate for difference *x*-coordinate, you can get the angular coefficient of the line:

**k****=**(Y2 - Y1) / (X2 - X1)- = (8-2)/(13-1)
- = 6/12
**= 1/2**

Since the slope is positive, the direction of the line is increasing. Since | k | **k****=** (Y2 - Y1) / (X2 - X1)

**= — 6**

Since the slope is negative, the straight line decreases. Since | k |> 1, the slope of the line is quite steep (slope> 45 °).

Synonyms:slope, tangent, slope of the line, slope

## The slope of the line and the slope of the line

Before writing such an equation, it is necessary to determine the angle of inclination of the line to the O x axis with their angular coefficient. Suppose that a Cartesian coordinate system O x on a plane is given.

**The angle of inclination of the line to the axis O x** located in the Cartesian coordinate system O x y on the plane, this is the angle that is measured from the positive direction O x to the line counterclockwise.

When the line is parallel to O x or there is a match in it, the angle of inclination is 0. Then the slope of the given line α is defined on the interval [0, π).

**Line angle coefficient** Is the tangent of the angle of the given line.

The standard designation is the letter k. From the definition we get that k = t g α. When the line is parallel to Oh, they say that the slope does not exist, since it goes to infinity.

The angular coefficient is positive when the graph of the function increases and vice versa. The figure shows various variations in the location of the right angle relative to the coordinate system with the coefficient value.

To find this angle, it is necessary to apply the definition of the angular coefficient and calculate the tangent of the angle of inclination in the plane.

Calculate the slope of a straight line at an angle of inclination of 120 °.

From the condition we have that α = 120 °. By definition, it is necessary to calculate the angular coefficient. We find it from the formula k = t g α = 120 = - 3.

If the angular coefficient is known, and it is necessary to find the angle of inclination to the abscissa axis, then the value of the angular coefficient should be taken into account. If k> 0, then the angle of the line is sharp and is found by the formula α = a r c t g k. If k 0, then the angle is obtuse, which gives the right to determine it by the formula α = π - a r c t g k.

Determine the angle of inclination of a given line to O x with an angular coefficient of 3.

From the condition we have that the slope is positive, which means that the angle of inclination to O x is less than 90 degrees. The calculations are made according to the formula α = a r c t g k = a r c t g 3.

**Answer: α = a r c t g 3.**

Find the angle of inclination of the line to the axis O x, if the angular coefficient = - 1 3.

If we take the letter k for the designation of the angular coefficient, then α is the angle of inclination to a given line in the positive direction О x. Hence k = - 1 3 0, then it is necessary to apply the formula α = π - a r c t g k When substituting, we obtain the expression:

α = π - a r c t g - 1 3 = π - a r c t g 1 3 = π - π 6 = 5 π 6.

**Answer:** 5 π 6 .

## Equation with an angular coefficient

An equation of the form y = k · x + b, where k is an angular coefficient and b is some real number, is called the equation of a line with an angular coefficient. The equation is characteristic of any straight, non-parallel axis O y.

If we consider in detail the straight line on the plane in a fixed coordinate system, which is given by an equation with an angular coefficient, which has the form y = k · x + b. In this case, it means that the coordinates of any point on the line correspond to the equation. If we substitute the coordinates of the point M, M 1 (x 1, y 1), in the equation y = k · x + b, then in this case the line will pass through this point, otherwise the point does not belong to the line.

A line with an angular coefficient y = 1 3 x - 1 is given. Calculate whether the points M 1 (3, 0) and M 2 (2, - 2) belong to a given line.

It is necessary to substitute the coordinates of the point M 1 (3, 0) in the given equation, then we get 0 = 1 3 · 3 - 1 ⇔ 0 = 0. Equality is true, then the point belongs to the line.

If we substitute the coordinates of the point M 2 (2, - 2), then we obtain an incorrect equality of the form - 2 = 1 3 · 2 - 1 ⇔ - 2 = - 1 3. We can conclude that the point M 2 does not belong to the line.

**Answer:** M 1 belongs to the line, but M 2 does not.

It is known that the line is defined by the equation y = k · x + b passing through M 1 (0, b), and when substituted, we obtain equality of the form b = k · 0 + b ⇔ b = b. From this we can conclude that the equation of a line with an angular coefficient y = k · x + b in the plane defines a line that passes through point 0, b. It forms an angle α with the positive direction of the axis O x, where k = t g α.

Consider, for example, a line defined using an angular coefficient defined by the form y = 3 · x - 1. We get that the line passes through the point with the coordinate 0, - 1 with a slope of α = a r c t g 3 = π 3 radians in the positive direction of the O x axis. This shows that the coefficient is 3.

## Equation of a line with an angular coefficient passing through a given point

It is necessary to solve the problem where it is necessary to obtain the equation of a line with a given angular coefficient passing through the point M 1 (x 1, y 1).

The equality y 1 = k · x + b can be considered fair, since the line passes through the point M 1 (x 1, y 1). To remove the number b, it is necessary to subtract the equation with an angular coefficient from the left and right sides. It follows that y - y 1 = k · (x - x 1). This equality is called the equation of a line with a given angular coefficient k passing through the coordinates of the point M 1 (x 1, y 1).

Make the equation of a line passing through point M 1 with coordinates (4, - 1), with an angular coefficient equal to - 2.

*Decision*

By hypothesis, we have that x 1 = 4, y 1 = - 1, k = - 2. Hence, the equation of the line is written in this way y - y 1 = k · (x - x 1) ⇔ y - (- 1) = - 2 · (x - 4) ⇔ y = - 2 x + 7.

**Answer:** y = - 2 x + 7.

Write the equation of a line with an angular coefficient that passes through point M 1 with coordinates (3, 5) parallel to the line y = 2 x - 2.

By hypothesis, we have that parallel lines have coincident angles of inclination, which means that the angular coefficients are equal. To find the angular coefficient from this equation, it is necessary to recall its basic formula y = 2 x - 2, which implies that k = 2. We compose an equation with an angular coefficient and we obtain:

y - y 1 = k · (x - x 1) ⇔ y - 5 = 2 · (x - 3) ⇔ y = 2 x - 1

## Transition from the equation of a line with an angular coefficient to other types of equations of a line and vice versa

Such an equation is not always applicable for solving problems, since it has a not very convenient record. To do this, it is necessary to present in a different form. For example, an equation of the form y = k · x + b does not allow writing down the coordinates of the directing vector of the line or the coordinates of the normal vector. To do this, you need to learn how to represent equations of a different kind.

We can obtain the canonical equation of a line on a plane using the equation of a line with an angular coefficient. We get x - x 1 a x = y - y 1 a y. It is necessary to transfer the term b to the left side and divide by the expression of the obtained inequality. Then we obtain an equation of the form y = k · x + b ⇔ y - b = k · x ⇔ k · x k = y - b k ⇔ x 1 = y - b k.

The equation of a line with an angular coefficient has become the canonical equation of this line.

Reduce the equation of a straight line with an angular coefficient y = - 3 x + 12 to the canonical form.

We calculate and represent the straight line in the form of the canonical equation. We get an equation of the form:

y = - 3 x + 12 ⇔ - 3 x = y - 12 ⇔ - 3 x - 3 = y - 12 - 3 ⇔ x 1 = y - 12 - 3

**Answer: x 1 = y - 12 - 3.**

The general equation of the line is easiest to obtain from y = k · x + b, but for this it is necessary to make the transformations: y = k · x + b ⇔ k · x - y + b = 0. A transition is made from the general equation of the line to equations of a different kind.

A direct equation of the form y = 1 7 x - 2 is given. Find out if a vector with coordinates a → = (- 1, 7) is a normal line vector?

To solve it, it is necessary to switch to another form of this equation, for this we write:

y = 1 7 x - 2 ⇔ 1 7 x - y - 2 = 0

The coefficients before the variables are the coordinates of the normal line vector. We write it like this n → = 1 7, - 1, hence 1 7 x - y - 2 = 0. It is clear that the vector a → = (- 1, 7) is collinear to the vector n → = 1 7, - 1, since we have a fair relation a → = - 7 · n →. It follows that the original vector a → = - 1, 7 is the normal vector of the line 1 7 x - y - 2 = 0, which means it is considered a normal vector for the line y = 1 7 x - 2.

We solve the problem inverse to this.

It is necessary to pass from the general form of the equation A x + B y + C = 0, where B ≠ 0, to the equation with an angular coefficient. for this we solve the equation with respect to y. We get A x + B y + C = 0 ⇔ - A B · x - C B.

The result is an equation with an angular coefficient that equals - A B.

A direct equation of the form 2 3 x - 4 y + 1 = 0 is given. Get the equation of this line with an angular coefficient.

Based on the condition, it is necessary to solve relative to y, then we obtain an equation of the form:

2 3 x - 4 y + 1 = 0 ⇔ 4 y = 2 3 x + 1 ⇔ y = 1 4 · 2 3 x + 1 ⇔ y = 1 6 x + 1 4.

**Answer: y = 1 6 x + 1 4.**

The equation of the form x a + y b = 1, which is called the equation of a straight line in segments, or the canonical form of the form x - x 1 a x = y - y 1 a y, is solved in a similar way. It is necessary to solve it with respect to y, only then we obtain an equation with an angular coefficient:

x a + y b = 1 ⇔ y b = 1 - x a ⇔ y = - b a · x + b.

The canonical equation can be reduced to a form with an angular coefficient. For this:

x - x 1 ax = y - y 1 ay ⇔ ay · (x - x 1) = ax · (y - y 1) ⇔ ⇔ ax · y = ay · x - ay · x 1 + ax · y 1 ⇔ y = ayax x - ayax x 1 + y 1

There is a line defined by the equation x 2 + y - 3 = 1. Bring equations with an angular coefficient to the form.

Based on the condition, it is necessary to transform, then we get an equation of the form _formula_. Both parts of the equation should be multiplied by - 3 in order to obtain the necessary equation with an angular coefficient. Transforming, we get:

y - 3 = 1 - x 2 ⇔ - 3 · y - 3 = - 3 · 1 - x 2 ⇔ y = 3 2 x - 3.

**Answer:** y = 3 2 x - 3.

Direct equation of the form x - 2 2 = y + 1 5 lead to a view with an angular coefficient.

It is necessary to calculate the expression x - 2 2 = y + 1 5 as a proportion. We get that 5 · (x - 2) = 2 · (y + 1). Now you need to completely resolve it, for this:

5 · (x - 2) = 2 · (y + 1) ⇔ 5 x - 10 = 2 y + 2 ⇔ 2 y = 5 x - 12 ⇔ y = 5 2 x

**Answer: y = 5 2 x - 6.**

To solve such problems, one should bring parametric equations of a straight line of the form x = x 1 + a x · λ y = y 1 + a y · λ to the canonical equation of the line, only after that one can proceed to the equation with an angular coefficient.

Find the slope of the line if it is given by the parametric equations x = λ y = - 1 + 2 · λ.

It is necessary to make the transition from a parametric view to an angular coefficient. To do this, we find the canonical equation from the given parametric:

x = λ y = - 1 + 2 · λ ⇔ λ = x λ = y + 1 2 ⇔ x 1 = y + 1 2.

Now it is necessary to resolve this equality with respect to y in order to obtain the equation of a line with an angular coefficient. for this we write in this way:

x 1 = y + 1 2 ⇔ 2 · x = 1 · (y + 1) ⇔ y = 2 x - 1

It follows that the slope of the straight line is 2. This is written as k = 2.