Coordinate Geometry of Lines

Coordinate geometry bridges algebra and geometry by using graphs and equations to describe lines. Two of the most important concepts are understanding how slopes relate to parallel and perpendicular lines, and calculating distances on the coordinate plane.

Slopes of Parallel and Perpendicular Lines

The slope of a line, often denoted by $m$, measures its steepness. By comparing the slopes of two lines, we can determine their geometric relationship:

• Parallel Lines: Two non-vertical lines are parallel if and only if they have the exact same slope. $m_1 = m_2$
• Perpendicular Lines: Two non-vertical lines are perpendicular (intersecting at a $90^\circ$ angle) if their slopes are negative reciprocals of each other. Their product is always $-1$. $m_1 \cdot m_2 = -1 \quad \text{or} \quad m_2 = -\frac{1}{m_1}$

Finding the Equation of a Line

If you know a point on a line $(x_1, y_1)$ and its slope $m$, you can find its equation using the point-slope form: $y - y_1 = m(x - x_1)$

Example: Find the equation of the line through $(2, 5)$ perpendicular to $y = 3x - 1$.

1. Identify the slope: The given line is $y = 3x - 1$, which is in slope-intercept form ($y = mx + b$). Its slope is $m_1 = 3$.
2. Find the perpendicular slope: The slope of our new line must be the negative reciprocal, so $m_2 = -\frac{1}{3}$.
3. Use the point-slope form: Substitute $m = -\frac{1}{3}$ and the point $(2, 5)$. $y - 5 = -\frac{1}{3}(x - 2)$
4. Simplify (optional): Multiply the entire equation by $3$ to clear the fraction. $3(y - 5) = -1(x - 2)$ $3y - 15 = -x + 2 \implies x + 3y - 17 = 0$

Distance from a Point to a Line

To find the shortest (perpendicular) distance $d$ from a point $(x_1, y_1)$ to a line given in the standard form $Ax + By + C = 0$, use the distance formula: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Example: Find the distance from point $(3, 4)$ to the line $2x - y + 1 = 0$.

1. Identify the components: From the line equation, $A = 2$, $B = -1$, and $C = 1$. From the point, $x_1 = 3$ and $y_1 = 4$.
2. Apply the formula: $d = \frac{|2(3) + (-1)(4) + 1|}{\sqrt{2^2 + (-1)^2}}$ $d = \frac{|6 - 4 + 1|}{\sqrt{4 + 1}}$ $d = \frac{|3|}{\sqrt{5}} = \frac{3}{\sqrt{5}}$
3. Rationalize the denominator: Multiply the top and bottom by $\sqrt{5}$. $d = \frac{3\sqrt{5}}{5}$