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Understanding the slope of a line is key when dealing with algebraic equations of lines. The slope tells us how steep a line is. If you think about a mountain slope, steepness determines how hard it is to climb the mountain. In algebra, the slope (\(m\)) of a line is the 'steepness' of that line on a graph. It's usually represented in the equation of a line as \(y = mx + c\).

Here:

• \(m\) is the slope.
• \(c\) is the y-intercept (where the line cuts the y-axis).

For example, in the line \(g(x) = -0.01x + 2.01\), the slope is \(-0.01\). This indicates a slightly descending line. A negative slope always indicates that as the x-value increases (moving right along the x-axis), the y-value decreases (it goes down).

It's important to see how slopes help describe the relationship between variables in linear equations.

The point-slope form of a line equation is a powerful tool in algebra when you know a point on the line and its slope. This form is written as: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope.

Using this formula lets you define a line by directly inserting the slope and a specific point.

• Substitute the given point into \((x_1, y_1)\).
• Replace \(m\) with the known slope.

For instance, if you have a slope of \(100\) and a point \((1, 2)\) through which the line passes, you can express the equation as \(y - 2 = 100(x - 1)\). This equation easily converts into other forms such as the slope-intercept form, making computation straightforward.

The point-slope form simplifies deriving an equation from specific line properties.

The slope-intercept form is often the most familiar linear equation format: \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.

This form clearly reflects both the slope and where the line crosses the y-axis, helping quickly sketch graphs or understand linear relationships.

• The slope (\(m\)) determines the line's angle. A higher absolute value means a steeper line.
• The y-intercept (\(c\)) is the starting value where x equals zero.

Using the problem's solution, we start with a point-slope form \(y - 2 = 100(x - 1)\) and simplify it:

• Distribute the \(100\): \(y - 2 = 100x - 100\).
• Add \(2\) to both sides to isolate \(y\): \(y = 100x - 98\).

This final equation shows a line with a steep positive slope that crosses the y-axis at \(-98\). It demonstrates how slope-intercept form supports understanding a line's graph and behavior.