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Chapter 2: Problem 95

Explain how to derive the slope-intercept form of a line's equation, \(y=m x+b,\) from the point-slope form \(y-y_{1}=m\left(x-x_{1}\right)\)

Short Answer

Expert verified
The equation can be transformed to the slope-intercept form through a two-step process: rearranging the equation \(y - y_1 = m(x - x_1)\) to \(y = m(x - x_1) + y_1\), and then simplifying it to \(y = mx - mx_1 + y_1 = mx + (-mx_1 + y_1)\), with \(-mx_1 + y_1\) equating to the parameter \(b\) in the slope-intercept form.

Step by step solution

01

Understanding Point-Slope Form

Firstly, understand that the point-slope form of a line's equation is given by \(y - y_1 = m(x - x_1)\). Here, \(m\) represents the slope of the line, while \((x_1, y_1)\) are the coordinates of a specific point on the line.
02

Rearrange the Equation

The next step is to rearrange the equation to the form \(y = mx + b\). This can be done by adding \(y_1\) to both sides of the equation. The resulting equation is \(y = m(x - x_1) + y_1\). This looks similar to the slope-intercept form, but there is still one step to completely transform it.
03

Simplify Equation to Slope-Intercept Form

To simplify it to the slope-intercept form, realize that the terms \(mx\) and \(-mx_1 + y_1\) are effectively in the \(y = mx + b\) form. Thus, identify \(b = -mx_1 + y_1\), with \(b\) representing the y-intercept (the y-value at which the line crosses the y-axis).

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