Rectangular Coordinates, Distance Formula, and Midpoint Formula Study Notes

Coordinate System Notation and Delta Operations

• Rectangular and Cartesian coordinates specify the location of points on a two-dimensional plane relative to a central origin point $(0, 0)$.

• Points are written as ordered pairs $(x, y)$, where $x$ represents horizontal position and $y$ represents vertical position.

• The Greek letter $\text{Delta}$ ($\text{Delta} = \frac{}{}$ or $\text{Delta} = \text{Change}$) is denoted by the symbol $\text{Delta}$:

  • $\text{Delta} = \text{Delta} = \text{Change}$

  • Change in the horizontal x-coordinate is represented as $\text{Delta} x = x_2 - x_1$

  • Change in the vertical y-coordinate is represented as $\text{Delta} y = y_2 - y_1$

• Specific context references include University, Mankato.

Pythagorean Theorem Principles and Verification

• The Pythagorean Theorem applies to right-angled triangles (RTA).

• Statement of the Theorem: In any triangle with side lengths $a$, $b$, and hypotenuse $c$ (where $c$ is the longest side located directly opposite the $90^\circ$ right angle):

  • $a^2 + b^2 = c^2$

• Example 1: Calculating Hypotenuse Length

  • Given side lengths $a = 3$ and $b = 4$:

  • $3^2 + 4^2 = c^2$

  • $9 + 16 = 25$

  • $c = \sqrt{25} = 5$ or $c = -5$

  • Because geometric side length and distance must be non-negative ($c \ge 0$), the negative solution is rejected, yielding $c = 5$

• Example 2: Testing Right-Angled Triangle (RTA) Status

  • Given side lengths $5$, $11$, and $13$ (with $13$ as the candidate hypotenuse):

  • $5^2 + 11^2 = 13^2$

  • $25 + 121 = 146$

  • $13^2 = 169$

  • Comparison: $146 \neq 169$

  • Conclusion: Side lengths $5$, $11$, and $13$ do not satisfy the theorem, meaning this triangle is Not RTA.

Mathematical Derivation of the Two-Dimensional Distance Formula

• Derivation of the distance $PQ$ between point $P(x_1, y_1)$ and point $Q(x_2, y_2)$:

  • Horizontal leg length: $\text{Delta} x = x_2 - x_1$

  • Vertical leg length: $\text{Delta} y = y_2 - y_1$

  • Applying the Pythagorean Theorem to the right triangle formed by these legs:

  • $(\text{Delta} x)^2 + (\text{Delta} y)^2 = (PQ)^2$

  • $(x_2 - x_1)^2 + (y_2 - y_1)^2 = (PQ)^2$

  • Taking the principal square root of both sides gives the Distance Formula:

  • $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = PQ$

• One-Dimensional Distance Example:

  • Distance between coordinates along an axis: $RQ = |-5 - (-1)| = 4$

Distance Calculations and Right-Angle Verification of Triangles

• Consideration of a triangle with vertices $A(1, 1)$, $B(3, 2)$, and $C(2, 4)$:

• Calculation of distance $AB$:

  • $AB = \sqrt{(3 - 1)^2 + (2 - 1)^2}$

  • $AB = \sqrt{2^2 + 1^2}$

  • $AB = \sqrt{4 + 1} = \sqrt{5}$

• Calculation of distance $AC$:

  • $AC = \sqrt{(2 - 1)^2 + (4 - 1)^2}$

  • $AC = \sqrt{1^2 + 3^2}$

  • $AC = \sqrt{1 + 9} = \sqrt{10}$

• Calculation of distance $BC$:

  • $BC = \sqrt{(2 - 3)^2 + (4 - 2)^2}$

  • $BC = \sqrt{(-1)^2 + 2^2}$

  • $BC = \sqrt{1 + 4} = \sqrt{5}$

• Verification of Right-Angled Triangle Status (RTA) for Triangle $ABC$:

  • Test if $AB^2 + BC^2 = AC^2$:

  • $(\sqrt{5})^2 + (\sqrt{5})^2 = (\sqrt{10})^2$

  • $5 + 5 = 10$

  • $10 = 10$

  • Conclusion: RTA = yes (Triangle $ABC$ is a right-angled triangle with hypotenuse $AC$).

Midpoint Formula and Segment Verification

• The Midpoint Formula determines the coordinates of the midpoint $M(x, y)$ midway between point $(x_1, y_1)$ and point $(x_2, y_2)$:

  • $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

• Example Calculation:

  • Point $P = (-1, 3)$

  • Point $Q = (1, 7)$

  • $M = \left(\frac{-1 + 1}{2}, \frac{3 + 7}{2}\right)$

  • $M = \left(\frac{0}{2}, \frac{10}{2}\right) = (0, 5)$

• Distance Verification from Midpoint $M(0, 5)$ to Endpoints:

  • Distance $PM$ from $P(-1, 3)$ to $M(0, 5)$:

  • $PM = \sqrt{(0 - (-1))^2 + (5 - 3)^2}$

  • $PM = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$

  • Distance $MQ$ from $M(0, 5)$ to $Q(1, 7)$:

  • $MQ = \sqrt{(1 - 0)^2 + (7 - 5)^2}$

  • $MQ = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$

  • Conclusion: $PM = MQ = \sqrt{5}$, confirming $M(0, 5)$ is equidistant from $P$ and $Q$.