91Ó°ÊÓ

The slope of a line is a way to describe how steep the line is. It tells us how much the line inclines or declines as we move from one point to another. Mathematically, if we have two points on a line, \( (x_1, y_1) \) and \( (x_2, y_2) \), then the slope can be calculated by the formula: \[m = \frac{y_2 - y_1}{x_2 - x_1}.\] For a line expressed in the standard form \( Ax + By + C = 0 \), you can rearrange it to the slope-intercept form \( y = mx + b \) to find the slope \( m \). In our example, the line was \( y - x + 3 = 0 \) which rearranges to \( y = x - 3 \), indicating the slope is 1. This means for each unit increase in \( x \), \( y \) increases by the same amount.

The equation of a line gives us a way to represent all points on the line in a simple equation. The most common form is the slope-intercept form: \( y = mx + b \). Here, \( m \) stands for the slope and \( b \) is the y-intercept—the point where the line crosses the y-axis. This form makes it easy to understand how the line is behaving on a graph.
In the exercise, we had a line with the equation \( y = x - 3 \), which tells us:

• The slope \( m \) is 1.
• The line crosses the y-axis at \( b = -3 \).

Another useful form is the point-slope formula: \( y - y_1 = m(x - x_1) \), which uses a known point \( (x_1, y_1) \) on the line. This is helpful when you have a specific point and the slope and want the equation of the line.

A normal line is a line perpendicular to a tangent line at a given point on a curve. In simple words, it is the line that forms a right angle with the tangent line. Calculating the slope of a normal line involves taking the negative reciprocal of the tangent line's slope.
For instance, if the slope of a given line is 1, then the slope of the normal line would be \( -1 \), since \[ m_{\text{normal}} = -\frac{1}{m_{\text{original}}}. \] In the exercise solution, we used this rule to find that the normal line to the given line at a specific point had a slope of \(-1\). This idea is vital in many mathematical fields, as normal lines can often represent the shortest path to a curve from a given point.

The center of a circle is a point within it, equidistant from all points on the circle itself. The center is represented in the general circle equation: \((x - h)^2 + (y - k)^2 = r^2\), where \( (h, k) \) denotes the center and \( r \) is the radius of the circle.
In the exercise, knowing where the center is helped determine which line is normal to the given circles.
Having the center on a normal line means \( (h, k) \) must satisfy the equation of the normal line.

• The given line was \( y = -x + \left(3 + 2\frac{3}{\sqrt{2}}\right) \).
• Checking centers allowed us to find that only the circle in option (A) matched.

Understanding this concept is essential when geometry problems involve circles and lines because it gives a definite point in space from which distances can be measured.