5 Best Ways to Check Whether the Point (x, y) Lies on a Given Line in Python

Rate this post

πŸ’‘ Problem Formulation: Determining if a specific point with coordinates (x, y) falls on a line defined by an equation or endpoints is a common computational geometry problem. In Python, this might involve input like a point (2, 3) and a line equation ‘y = 2x + 1’ and checking whether the point satisfies the line’s equation, which in this case, it doesn’t, as the desired output is False.

Method 1: Algebraic Approach

An algebraic approach involves substituting the point (x, y) into the line equation and checking if the equation holds true. If the line equation is ‘y = mx + c’ and the point is (x, y), the method checks whether the left side of the equation equals the right side, when x and y values are substituted in.

Here’s an example:

def is_on_line(x, y, m, c):
    return y == m * x + c

# Example usage:
point_x, point_y = (2, 3)
slope_m = 2
y_intercept_c = 1

print(is_on_line(point_x, point_y, slope_m, y_intercept_c))

Output: False

This code portrays an algebraic approach where the ‘is_on_line’ function checks if the inputted point (2, 3) is on the line with the slope of 2 and y-intercept of 1, by verifying if substituting x and y into ‘y = mx + c’ holds true. It doesn’t, hence returning False.

Method 2: Vector Cross Product

Using cross product of vectors, we can determine whether a point is on a straight line by comparing it with two known points on the line. This method is suitable when a line is defined by two points rather than a slope-intercept form.

Here’s an example:

def is_on_line_by_vectors(x1, y1, x2, y2, px, py):
    return (x2 - x1) * (py - y1) == (y2 - y1) * (px - x1)

# Example usage:
point1 = (1, 2)
point2 = (3, 6)
point_x, point_y = (2, 3)

print(is_on_line_by_vectors(*point1, *point2, point_x, point_y))

Output: False

The function ‘is_on_line_by_vectors’ checks if the difference in coordinates of the known line points and the point in question results in a cross product equal to zero, indicating collinearity and hence, the point (2, 3) lying on the line. It evaluates to False, showing the point is not on the line defined by points (1, 2) and (3, 6).

Method 3: Geometrical Slope Comparison

This method compares the slope between two known points on the line with the slope between one of those points and the query point. If both slopes are equal, the query point lies on the line.

Here’s an example:

def is_on_line_by_slope(p1, p2, q):
    (x1, y1), (x2, y2), (qx, qy) = p1, p2, q
    return (y2 - y1) * (qx - x1) == (qy - y1) * (x2 - x1)

# Example usage:
point1 = (1, 2)
point2 = (3, 6)
query_point = (2, 3)

print(is_on_line_by_slope(point1, point2, query_point))

Output: False

The ‘is_on_line_by_slope’ function computes if the slopes match by checking if the cross-multiplication of the differences in coordinates is equal, indicative of equal slopes and thus the point (2, 3) being on the line. The check here is false, indicating the point in question does not lie on the line.

Method 4: Using NumPy Library

NumPy, a popular Python library for numerical calculations, can be used to solve this problem efficiently. NumPy’s array operations can simplify checking if a point lies on a given line by applying vectorized computations.

Here’s an example:

import numpy as np

def is_on_line_numpy(point, line_coefficients):
    x, y = point
    m, c = line_coefficients
    return np.isclose(y, m * x + c)

# Example usage:
point = np.array([2, 3])
line_coeffs = np.array([2, 1])

print(is_on_line_numpy(point, line_coeffs))

Output: False

The ‘is_on_line_numpy’ function leverages NumPy’s ‘isclose’ method to account for floating-point precision issues when checking if the point satisfies the line equation ‘y = mx + c’. This method is significantly faster for large datasets but introduces a dependency on an external library.

Bonus One-Liner Method 5: Using a Lambda Function

A lambda function provides a quick, one-liner method in Python for checks that can be expressed as a single expression, like the point-on-line check.

Here’s an example:

is_on_line = lambda x, y, m, c: y == m * x + c

# Example usage:
print(is_on_line(2, 3, 2, 1))

Output: False

The one-liner uses a lambda function to encapsulate the algebraic check into a concise expression that reads almost like mathematical notation. It checks if point (2, 3) lies on the line ‘y = 2x + 1’. This method is elegant and straightforward but lacks the explicitness and clarity that a full function definition offers.

Summary/Discussion

  • Method 1: Algebraic Approach. Direct and straightforward. Limited to lines expressed in slope-intercept form.
  • Method 2: Vector Cross Product. Does not require the line to be in slope-intercept form. Involves more computational steps.
  • Method 3: Geometrical Slope Comparison. Relies on comparing slopes. Simple, but calculating slopes could be redundant if line equation is known.
  • Method 4: Using NumPy Library. Efficient and robust against floating-point issues. Requires NumPy installed, adding overhead for small problems.
  • Bonus Method 5: Using a Lambda Function. Concise and quick for simple checks. Might sacrifice code readability and is less descriptive than other methods.