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The slope-intercept form is a widely used way to express the equation of a straight line. It is generally written as \( y = mx + c \), where \( m \) is the slope of the line and \( c \) is the y-intercept, the point where the line crosses the y-axis.

Understanding the slope \( m \) is crucial. It indicates how steep the line is. If the slope is positive, the line goes upwards as it moves from left to right. If negative, it goes downwards. A slope of zero means the line is horizontal. In our exercise, we have a slope \( m = 4 \), showing a fairly steep upward line.

Recall that the y-intercept \( c \) tells you where the line crosses the y-axis. For example, if \( c = -1 \), the line will intersect the y-axis at -1. Finding \( c \) generally involves substituting the x and y coordinates of a point the line passes through into the equation followed by solving for \( c \). Starting with \( m = 4 \) and knowing the line goes through the point \((1,3)\), we determine \( c \) by plugging \( x = 1 \) and \( y = 3 \) into the equation \( 3 = 4 \times 1 + c \), simplifying to \( c = -1 \).

And there you have it! The equation \( y = 4x - 1 \) is in slope-intercept form, poised for conversion into other forms if needed.

The standard form of a linear equation is expressed as \( Ax + By = C \). This presentation can be particularly useful for solving systems of equations or finding x and y intercepts easily. Important to note, \( A \), \( B \), and \( C \) are integers, and \( A \) should be non-negative.

To transition our equation from slope-intercept form \( y = 4x - 1 \) to standard form, we aim to isolate zero on one side, moving terms around if necessary. Here's how it's done: subtract \( y \) from both sides, yielding \( 4x - y = 1 \). The terms \( A = 4 \), \( B = -1 \), and \( C = 1 \) neatly align with the requirements of the standard form. Here, the entire expression is set to equal a constant (\(1\)), a defining characteristic of standard form.

This rearrangement makes it easier to interpret certain characteristics of the equations and can additionally streamline solving complex algebraic problems.

The point-slope equation is a powerful tool that uses a known point on a line along with the slope to construct a line's equation. The formula is given by \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is a known point on the line and \( m \) is the slope.

For example, given our slope \( m = 4 \) and a point \((1, 3)\), we can substitute these values into the point-slope formula:\( y - 3 = 4(x - 1) \).

Here's how it works:

• Start with the equation \( y - 3 = 4(x - 1) \).
• Distribute the slope \( 4 \) across the \( (x - 1) \): \( y - 3 = 4x - 4 \).
• Add \( 3 \) to both sides to isolate \( y \): \( y = 4x - 1 \).

This will confirm the equation derived earlier, \( y = 4x - 1 \). The point-slope form provides a direct bridge to the slope-intercept form, confirming that understanding these conversion techniques can enhance versatility in solving various linear problems.