Why are A B Coordinates of the Normal Vector of the Line AxByC0?

The equation of a line in the form Ax By C 0 is a fundamental representation in two-dimensional Cartesian coordinate systems. Analyzing this equation helps in understanding the properties of its normal vector, which is a vector that is perpendicular to the line. In this article, we delve into the relationship between the coefficients A and B and the normal vector of the line.

Standard Form of a Line

The equation Ax By C 0 describes a straight line in a two-dimensional Cartesian coordinate system. It is a linear equation where A, B, and C are constants, and x and y are variables. This form allows us to represent any line in the plane, providing a clear mathematical description of its orientation and position.

Identifying the Normal Vector

A normal vector to a line is a vector that is perpendicular to the line. This vector is crucial for understanding the orientation of the line and is extensively used in various mathematical and engineering applications. Let's explore how to identify the normal vector for the line represented by the equation Ax By C 0.

Rearranging the Equation

To express the line equation in a standard form, we can rearrange it as follows:

By -Ax - C rarr; y -frac{A}{B}x - frac{C}{B}

This algebraic manipulation transforms the line equation into a familiar slope-intercept form y mx b, where y is the dependent variable and x is the independent variable. From this, we can directly identify the slope of the line as -frac{A}{B}.

Perpendicular Slope

The slope of any line perpendicular to this one, which represents the direction of the normal vector, is the negative reciprocal of -frac{A}{B}. Therefore, the slope of the normal vector is:

frac{B}{A}

Normal Vector Components

The normal vector can be represented by a vector whose components correspond to the coefficients A and B in the original line equation. Specifically:

mathbf{n} begin{pmatrix} A B end{pmatrix}

In this representation, A and B are the coefficients from the line equation, directly reflecting the orientation of the line. Hence, the coordinates (A, B) of the normal vector correspond to the coefficients of x and y in the line equation Ax By C 0. This relationship clearly shows that the normal vector directly reflects the direction and orientation of the line in the coordinate plane.

Alternative Approach: Understanding through Two Points and Direction Vector

To further reinforce our understanding, let's consider an alternative method using two points P, Q and a direction vector (mathbf{u} overrightarrow{Q - P}).

Using Point and Direction Vector

Point M (x, y) on the line can be expressed as:

M P t . mathbf{u}

In coordinates, this becomes:

x P_x t . u_x

y P_y t . u_y

Solving for t in each equation:

t frac{x - P_x}{u_x}

t frac{y - P_y}{u_y}

By equating the two expressions for t:

u_y (x - P_x) - u_x (y - P_y) 0

or equivalently,

A x B y C 0

Here, A -u_y, B u_x, C u_y P_x - u_x P_y. This indicates that the vector (mathbf{n} (A, B)) is indeed the normal vector since it is perpendicular to the direction vector (mathbf{u}).

Conclusion

The coordinates (A, B) of the normal vector correspond directly to the coefficients of x and y in the line equation Ax By C 0. This relationship highlights the intrinsic connection between the line's equation and its normal vector, providing a clear and concise method for understanding the orientation and properties of lines in the Cartesian plane.