9.1 Distance, Midpoint, Slope Criteria & Coordinate Proofs

9.1 Distance, Midpoint, Slope Criteria & Coordinate Proofs

Foundational Metric Formulas: Euclidean Distance and Midpoint

The Cartesian coordinate system provides an algebraic bridge to Euclidean geometry, translating geometric postulates into analytic equations. The fundamental metric on $\mathbb{R}^2$ is the Euclidean distance between two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$. Constructing a right triangle with legs parallel to the coordinate axes yields horizontal displacement $\Delta x = x_2 - x_1$ and vertical displacement $\Delta y = y_2 - y_1$. Applying the Pythagorean theorem directly derives the Euclidean distance formula: $d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Because squaring eliminates sign dependency, $(x_2 - x_1)^2 = (x_1 - x_2)^2$, making distance invariant under endpoint ordering.

The midpoint $M$ of directed segment $\overline{P_1 P_2}$ represents the point equidistant from $P_1$ and $P_2$ that lies on the segment. Analytically, each coordinate of $M$ is the arithmetic mean of the corresponding endpoint coordinates: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ This formula generalizes to $n$-dimensional Euclidean space $\mathbb{R}^n$, reflecting the linear center of mass of two unit point masses.

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Directed Segment Partitioning in a Given Ratio

A recurring competency on the FTCE Mathematics 6-12 examination involves partitioning a directed line segment $\overline{AB}$ in a specified ratio. Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be distinct points on the Cartesian plane. We seek a point $P(x, y)$ on $\overline{AB}$ such that the ratio of the directed distance from $A$ to $P$ to the directed distance from $P$ to $B$ satisfies: $\frac{AP}{PB} = \frac{m}{n}, \quad \text{where } m, n > 0$ This is equivalent to stating that $P$ lies a fraction $k = \frac{m}{m + n}$ of the total directed distance from $A$ to $B$. By dropping perpendiculars to the coordinate axes, similar right triangles establish that the horizontal and vertical displacements scale by the identical ratio $k$: $x = x_1 + \frac{m}{m + n}(x_2 - x_1) = \frac{n x_1 + m x_2}{m + n}$ $y = y_1 + \frac{m}{m + n}(y_2 - y_1) = \frac{n y_1 + m y_2}{m + n}$ When $m = n = 1$, the partition ratio is $1:1$, reducing to the standard midpoint formula $k = \frac{1}{2}$. Examinees must carefully distinguish between two common problem phrasings:

1. "Point $P$ divides segment $\overline{AB}$ in the ratio $m:n$": Here, $k = \frac{m}{m + n}$.
2. "Point $P$ is $\frac{m}{n}$ of the distance from $A$ to $B$": Here, the fractional scalar $k = \frac{m}{n}$ is applied directly without adding the denominator.

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Slope Criteria, Point-to-Line Distance & Parallel Lines

The orientation of lines in the coordinate plane is quantified by slope $m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$ for non-vertical lines ($\Delta x \neq 0$). Geometric relationships translate into algebraic criteria:

• Parallel Lines: Two distinct non-vertical lines $L_1$ and $L_2$ with slopes $m_1$ and $m_2$ are parallel ($L_1 \parallel L_2$) if and only if their slopes are identical: $m_1 = m_2$.
• Perpendicular Lines: Two non-vertical lines $L_1$ and $L_2$ are perpendicular ($L_1 \perp L_2$) if and only if the product of their slopes is $-1$: $m_1 \cdot m_2 = -1 \iff m_2 = -\frac{1}{m_1}$ This negative reciprocal relationship arises directly from rotating a direction vector $(1, m_1)$ by $90^\circ$ to $(-m_1, 1)$, whose slope is $-1/m_1$.

The perpendicular distance from an arbitrary point $P(x_0, y_0)$ to a general line $L: Ax + By + C = 0$ is obtained by projecting the vector from a line point to $P$ onto the normal vector $\mathbf{n} = \langle A, B \rangle$: $d(P, L) = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}}$ For two parallel lines written in standard form with identical leading coefficients, $L_1: Ax + By + C_1 = 0$ and $L_2: Ax + By + C_2 = 0$, the perpendicular distance separating them is: $d(L_1, L_2) = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$ If given lines have proportional coefficients, one must multiply or divide by a scalar to equate $A$ and $B$ before applying this formula.

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Coordinate Proof Methodology: Strategic Variable Assignment

A coordinate proof uses algebra to verify geometric theorems. The power of coordinate proofs relies on setting up a coordinate system that maximizes computational simplicity without sacrificing geometric generality.

Key strategic placement heuristics include:

1. Anchor to the Origin: Place one vertex or significant point (e.g., center of symmetry or intersection of diagonals) at $(0, 0)$.
2. Align with Coordinate Axes: Position at least one side along the x-axis or y-axis so that one coordinate becomes identically zero.
3. Exploit Symmetry: For symmetric figures (isosceles triangles, rectangles, rhombi, isosceles trapezoids), place the axis of symmetry along the y-axis or place the figure symmetrically about the origin.
4. Avoid Fractions via Scale Multipliers: If midpoints will be computed, assign coordinates using multiples of 2 (e.g., $(2a, 0)$, $(2b, 2c)$) so that dividing by 2 yields clean integer expressions without fractions.
5. Preserve Generality: Never use numerical constants for general proofs. Avoid assigning special properties (e.g., do not set $b = a$ or $y = 0$ unless the geometric definition requires it).

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Worked Coordinate Proof Exemplars

Exemplar 1: Diagonals of a Parallelogram Bisect Each Other

• Figure Setup: Place parallelogram $ABCD$ with vertex $A$ at the origin $(0, 0)$. Align base $\overline{AB}$ along the positive x-axis by defining $B(a, 0)$ where $a > 0$. Let $D$ be located at $(b, c)$ with $c > 0$. Since $\overline{DC}$ is parallel and congruent to $\overline{AB}$, the vector from $D$ to $C$ is $\langle a, 0 \rangle$. Thus, vertex $C$ is $(b + a, c) = (a + b, c)$.
• Diagonal 1 ($\overline{AC}$): Connects $(0, 0)$ to $(a + b, c)$. Its midpoint $M_{AC}$ is: $M_{AC} = \left(\frac{0 + (a + b)}{2}, \frac{0 + c}{2}\right) = \left(\frac{a + b}{2}, \frac{c}{2}\right)$
• Diagonal 2 ($\overline{BD}$): Connects $(a, 0)$ to $(b, c)$. Its midpoint $M_{BD}$ is: $M_{BD} = \left(\frac{a + b}{2}, \frac{0 + c}{2}\right) = \left(\frac{a + b}{2}, \frac{c}{2}\right)$
• Conclusion: Because $M_{AC} = M_{BD}$, both diagonals intersect at their mutual midpoint, proving analytically that the diagonals of any parallelogram bisect each other.

Exemplar 2: Triangle Midsegment Theorem

• Figure Setup: Place $\triangle ABC$ on the plane with $A$ at $(0, 0)$, $B$ along the x-axis at $(2a, 0)$, and $C$ at $(2b, 2c)$. Using coefficients of 2 prevents fractional denominators.
• Midpoints: Let $M$ be the midpoint of $\overline{AC}$ and $N$ be the midpoint of $\overline{BC}$: $M = \left(\frac{0 + 2b}{2}, \frac{0 + 2c}{2}\right) = (b, c)$ $N = \left(\frac{2a + 2b}{2}, \frac{0 + 2c}{2}\right) = (a + b, c)$
• Parallelism Verification: Calculate the slopes of midsegment $\overline{MN}$ and base $\overline{AB}$: $m_{MN} = \frac{c - c}{(a + b) - b} = \frac{0}{a} = 0, \quad m_{AB} = \frac{0 - 0}{2a - 0} = 0$ Since $m_{MN} = m_{AB} = 0$, $\overline{MN} \parallel \overline{AB}$.
• Length Verification: Calculate lengths using the distance formula: $MN = \sqrt{((a + b) - b)^2 + (c - c)^2} = \sqrt{a^2 + 0} = |a| = a$ $AB = \sqrt{(2a - 0)^2 + (0 - 0)^2} = \sqrt{4a^2} = 2a$ Since $MN = a = \frac{1}{2}(2a) = \frac{1}{2}AB$, the midsegment is parallel to the third side and half its length.

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Worked Computational Exemplar: Directed Segment Partition

Problem: Find the coordinates of point $P$ that partitions directed segment $\overline{AB}$ from $A(-3, 8)$ to $B(7, -7)$ in the ratio $2:3$.

• Solution: The partition ratio is $m:n = 2:3$.
• The fractional distance along the segment from $A$ to $B$ is: $k = \frac{m}{m + n} = \frac{2}{2 + 3} = \frac{2}{5}$
• Apply the horizontal coordinate formula: $x_P = x_1 + k(x_2 - x_1) = -3 + \frac{2}{5}(7 - (-3)) = -3 + \frac{2}{5}(10) = -3 + 4 = 1$
• Apply the vertical coordinate formula: $y_P = y_1 + k(y_2 - y_1) = 8 + \frac{2}{5}(-7 - 8) = 8 + \frac{2}{5}(-15) = 8 - 6 = 2$
• Therefore, the partition point is $P(1, 2)$. Checking the directed distances confirms: $AP = \sqrt{(1 - (-3))^2 + (2 - 8)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$ $PB = \sqrt{(7 - 1)^2 + (-7 - 2)^2} = \sqrt{36 + 81} = \sqrt{117} = 3\sqrt{13}$ $\frac{AP}{PB} = \frac{2\sqrt{13}}{3\sqrt{13}} = \frac{2}{3}$

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Coordinate Proof Setup and Formula Reference Table