Slope of a Straight Line Calculator

Enter the coordinates of two points to get information about the line that passes through them (equations, y-intercept, angle of inclination, etc.), the step-by-step solution, and the graph.

Point 1
( , )
Point 2
( , )

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Solved Exercises

Calculate the slope and elements of the line that passes through the points (-2, 3) and (4, 7).

Result

The line that passes through the given points has the following characteristics, equations, and elements.

Slope: \( m = \dfrac{2}{3} \approx 0.67 \)

Slope-intercept form: \( y = \dfrac{2}{3}x + \dfrac{13}{3} \)

Point-slope form: \( y - 3 = \dfrac{2}{3}\left(x + 2\right) \)

General form: \( 2x - 3y + 13 = 0 \)

y-intercept: \( b = \dfrac{13}{3} \approx 4.33 \)

Angle of inclination: \( \theta \approx 33.69^\circ \)

Vertical change: \( \Delta y = 4 \)

Horizontal change: \( \Delta x = 6 \)

Distance to the origin: \( d \approx 3.61 \)

Slope in percentage: \( m_{\%} \approx 66.67\% \)

Step-by-step solution

1. Calculate the slope (m).

We use the slope formula for two points in a plane, which is the division between the vertical change (Δy) and the horizontal change (Δx):

$$ m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2 - y_1}{x_2 - x_1} $$

We substitute the coordinates of \( P_1(-2, 3) \), \( P_2(4, 7) \):

$$ m = \dfrac{7 - 3}{4 - \left(-2\right)} = \dfrac{4}{6} = \dfrac{2}{3} $$

2. Calculate the angle of inclination (θ).

The angle of inclination is obtained by applying the arctangent function to the calculated slope:

$$ \theta = \arctan(m) = \arctan\left(\dfrac{2}{3}\right) \approx 33.69^\circ $$

3. Find the y-intercept (b) and the equations of the line.

We start from the point-slope equation using the coordinates of P₁ (the result would be the same using P₂):

$$ y - y_1 = m(x - x_1) $$
$$ y - 3 = \dfrac{2}{3}\left(x - \left(-2\right)\right) $$
$$ y - 3 = \dfrac{2}{3}\left(x + 2\right) $$

We solve for the variable 'y' and expand to get the equation in slope-intercept form (y = mx + b):

$$ y = \dfrac{2}{3}x + \dfrac{13}{3} $$

From this equation we extract that the y-intercept is \( b = \dfrac{13}{3} \).

Starting from the slope-intercept form, we rearrange the terms to equal zero in the general form Ax + By + C = 0:

$$ 2x - 3y + 13 = 0 $$
Graph on the Cartesian plane of a line with a positive slope obtained from two points.
Graph of the line
Find the slope of the straight line where (1, 5) and (6, -2) are two of its points.

Answer

Slope: \( m = -\dfrac{7}{5} = -1.4 \)

Slope-intercept form: \( y = -\dfrac{7}{5}x + \dfrac{32}{5} \)

Point-slope form: \( y - 5 = -\dfrac{7}{5}\left(x - 1\right) \)

General form: \( 7x + 5y - 32 = 0 \)

y-intercept: \( b = \dfrac{32}{5} = 6.4 \)

Angle of inclination: \( \theta \approx 125.54^\circ \)

Vertical change: \( \Delta y = -7 \)

Horizontal change: \( \Delta x = 5 \)

Distance to the origin: \( d \approx 3.72 \)

Slope in percentage: \( m_{\%} \approx -140\% \)

Step-by-step solution

1. Calculate the slope (m).

We use the slope formula for two points in a plane, which is the division between the vertical change (Δy) and the horizontal change (Δx):

$$ m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2 - y_1}{x_2 - x_1} $$

We substitute the coordinates of \( P_1(1, 5) \), \( P_2(6, -2) \):

$$ m = \dfrac{-2 - 5}{6 - 1} = \dfrac{-7}{5} = -\dfrac{7}{5} $$

2. Calculate the angle of inclination (θ).

The angle of inclination is obtained by applying the arctangent function to the calculated slope:

$$ \theta = \arctan(m) = \arctan\left(-\dfrac{7}{5}\right) \approx -54.46^\circ $$

Since the slope is negative, the arctangent function returns a negative angle (measured clockwise). By geometric convention, the angle of inclination of a line is always expressed as a positive value between 0° and 180°.

To get the standard angle of inclination, we add 180° to the arctangent:

$$ \theta = \arctan\left(-\dfrac{7}{5}\right) + 180^\circ \approx 125.54^\circ $$

3. Find the y-intercept (b) and the equations of the line.

We start from the point-slope equation using the coordinates of P₁ (the result would be the same using P₂):

$$ y - y_1 = m(x - x_1) $$
$$ y - 5 = -\dfrac{7}{5}\left(x - 1\right) $$

We solve for the variable 'y' and expand to get the equation in slope-intercept form (y = mx + b):

$$ y = -\dfrac{7}{5}x + \dfrac{32}{5} $$

From this equation we extract that the y-intercept is \( b = \dfrac{32}{5} \).

Starting from the slope-intercept form, we rearrange the terms to equal zero in the general form Ax + By + C = 0:

$$ 7x + 5y - 32 = 0 $$
Graph on the Cartesian plane of a line with a negative slope obtained from two points.
Graph of the line
Determine the slope of the line given the points (-2, -2) and (3, 2).

Solution

Slope: \( m = \dfrac{4}{5} = 0.8 \)

Slope-intercept form: \( y = \dfrac{4}{5}x - \dfrac{2}{5} \)

Point-slope form: \( y + 2 = \dfrac{4}{5}\left(x + 2\right) \)

General form: \( 4x - 5y - 2 = 0 \)

y-intercept: \( b = -\dfrac{2}{5} = -0.4 \)

Angle of inclination: \( \theta \approx 38.66^\circ \)

Vertical change: \( \Delta y = 4 \)

Horizontal change: \( \Delta x = 5 \)

Distance to the origin: \( d \approx 0.31 \)

Slope in percentage: \( m_{\%} \approx 80\% \)

Step-by-step solution

1. Calculate the slope (m).

We use the slope formula for two points in a plane, which is the division between the vertical change (Δy) and the horizontal change (Δx):

$$ m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2 - y_1}{x_2 - x_1} $$

We substitute the coordinates of \( P_1(-2, -2) \), \( P_2(3, 2) \):

$$ m = \dfrac{2 - \left(-2\right)}{3 - \left(-2\right)} = \dfrac{4}{5} $$

2. Calculate the angle of inclination (θ).

The angle of inclination is obtained by applying the arctangent function to the calculated slope:

$$ \theta = \arctan(m) = \arctan\left(\dfrac{4}{5}\right) \approx 38.66^\circ $$

3. Find the y-intercept (b) and the equations of the line.

We start from the point-slope equation using the coordinates of P₁ (the result would be the same using P₂):

$$ y - y_1 = m(x - x_1) $$
$$ y - \left(-2\right) = \dfrac{4}{5}\left(x - \left(-2\right)\right) $$
$$ y + 2 = \dfrac{4}{5}\left(x + 2\right) $$

We solve for the variable 'y' and expand to get the equation in slope-intercept form (y = mx + b):

$$ y = \dfrac{4}{5}x - \dfrac{2}{5} $$

From this equation we extract that the y-intercept is \( b = -\dfrac{2}{5} \).

Starting from the slope-intercept form, we rearrange the terms to equal zero in the general form Ax + By + C = 0:

$$ 4x - 5y - 2 = 0 $$
Graph on the Cartesian plane of a line with a positive slope obtained from two points with integer coordinates.
Graph of the line on the plane
Obtain the elements of the straight line containing the points (-1/2, 3/4) and (5/2, -1/4).

Result

Slope: \( m = -\dfrac{1}{3} \approx -0.33 \)

Slope-intercept form: \( y = -\dfrac{1}{3}x + \dfrac{7}{12} \)

Point-slope form: \( y - \dfrac{3}{4} = -\dfrac{1}{3}\left(x + \dfrac{1}{2}\right) \)

General form: \( 4x + 12y - 7 = 0 \)

y-intercept: \( b = \dfrac{7}{12} \approx 0.58 \)

Angle of inclination: \( \theta \approx 161.57^\circ \)

Vertical change: \( \Delta y = -1 \)

Horizontal change: \( \Delta x = 3 \)

Distance to the origin: \( d \approx 0.55 \)

Slope in percentage: \( m_{\%} \approx -33.33\% \)

Step-by-step solution

1. Calculate the slope (m).

We use the slope formula for two points in a plane, which is the division between the vertical change (Δy) and the horizontal change (Δx):

$$ m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2 - y_1}{x_2 - x_1} $$

We substitute the coordinates of \( P_1(-\dfrac{1}{2}, \dfrac{3}{4}) \), \( P_2(\dfrac{5}{2}, -\dfrac{1}{4}) \):

$$ m = \dfrac{-\dfrac{1}{4} - \dfrac{3}{4}}{\dfrac{5}{2} - \left(-\dfrac{1}{2}\right)} = \dfrac{-1}{3} = -\dfrac{1}{3} $$

2. Calculate the angle of inclination (θ).

The angle of inclination is obtained by applying the arctangent function to the calculated slope:

$$ \theta = \arctan(m) = \arctan\left(-\dfrac{1}{3}\right) \approx -18.43^\circ $$

Since the slope is negative, the arctangent function returns a negative angle (measured clockwise). By geometric convention, the angle of inclination of a line is always expressed as a positive value between 0° and 180°.

To get the standard angle of inclination, we add 180° to the arctangent:

$$ \theta = \arctan\left(-\dfrac{1}{3}\right) + 180^\circ \approx 161.57^\circ $$

3. Find the y-intercept (b) and the equations of the line.

We start from the point-slope equation using the coordinates of P₁ (the result would be the same using P₂):

$$ y - y_1 = m(x - x_1) $$
$$ y - \dfrac{3}{4} = -\dfrac{1}{3}\left(x - \left(-\dfrac{1}{2}\right)\right) $$
$$ y - \dfrac{3}{4} = -\dfrac{1}{3}\left(x + \dfrac{1}{2}\right) $$

We solve for the variable 'y' and expand to get the equation in slope-intercept form (y = mx + b):

$$ y = -\dfrac{1}{3}x + \dfrac{7}{12} $$

From this equation we extract that the y-intercept is \( b = \dfrac{7}{12} \).

Starting from the slope-intercept form, we rearrange the terms to equal zero in the general form Ax + By + C = 0:

$$ 4x + 12y - 7 = 0 $$
Graph on the Cartesian plane of a line with a negative slope obtained from two points with fractional coordinates.
Graph of the line on the plane

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Daniel Machado

Professor of Mathematics, graduated from the Faculty of Exact, Chemical and Natural Sciences of the National University of Misiones (UNAM). Developer and creator of RigelUp, dedicated to building tools for mathematical learning.

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