Coordinate geometry (distance, midpoint, slopes) Knowledge-Based Study

Study notes

1. In coordinate geometry, the slope of a line measures its steepness and direction. Given two distinct points $A(x_1, y_1)$ and $B(x_2, y_2)$ where $x_1 \neq x_2$, the slope $m$ is the ratio of vertical change (rise) to horizontal change (run):

   $$m = \frac{y_2 - y_1}{x_2 - x_1}$$

   Key Characteristics of Slope:

   • Positive slope ($m > 0$): The line rises from left to right.
   • Negative slope ($m < 0$): The line falls from left to right.
   • Zero slope ($m = 0$): The line is completely horizontal because $y_2 - y_1 = 0$.
   • Undefined slope: The line is vertical because $x_2 - x_1 = 0$, and division by zero is undefined.

   Parallel and Perpendicular Lines:

   Let two non-vertical lines have slopes $m_1$ and $m_2$:

   • Parallel lines: Slopes are equal ($m_1 = m_2$).
   • Perpendicular lines: Slopes are negative reciprocals ($m_1 \cdot m_2 = -1$, or $m_2 = -\frac{1}{m_1}$).

   The midpoint of a line segment is the exact point that divides the segment into two congruent halves. Conceptually, it is the average of the coordinates of the endpoints.

   For any line segment with endpoints $A(x_1, y_1)$ and $B(x_2, y_2)$, the midpoint $M$ is given by:

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

   Worked Example:

   Find the midpoint of the segment connecting $P(-4, 7)$ and $Q(6, 1)$.

   • $x$-coordinate: $\frac{-4 + 6}{2} = \frac{2}{2} = 1$
   • $y$-coordinate: $\frac{7 + 1}{2} = \frac{8}{2} = 4$

   The coordinates of the midpoint are $M(1, 4)$.

   The distance formula calculates the straight-line distance between any two points in the Cartesian plane. It is derived directly from the Pythagorean Theorem.

   Consider two points $A(x_1, y_1)$ and $B(x_2, y_2)$. By drawing a horizontal line through $A$ and a vertical line through $B$, a right triangle is formed where:

   • The horizontal leg has length $|x_2 - x_1|$.
   • The vertical leg has length $|y_2 - y_1|$.
   • The hypotenuse $d$ is the distance between $A$ and $B$.

   Applying $a^2 + b^2 = c^2$:

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

   Worked Example:

   Find the distance between $A(2, -3)$ and $B(-1, 1)$:

   $$d = \sqrt{((-1) - 2)^2 + (1 - (-3))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

   Coordinate geometry allows us to rigorously verify the properties of geometric shapes by combining distance, midpoint, and slope:

   • Distance: Confirms congruence of sides (e.g., equilateral, isosceles, or scalene triangles).
   • Slope: Confirms right angles (perpendicular sides) and parallelism (opposite sides of parallelograms).
   • Midpoint: Confirms whether diagonals bisect each other.

   Worked Example:

   Verify whether triangle $ABC$ with vertices $A(2, 1)$, $B(5, 2)$, and $C(1, 4)$ is a right-angled triangle.

   • Slope of segment $AB$: $m_{AB} = \frac{2 - 1}{5 - 2} = \frac{1}{3}$
   • Slope of segment $AC$: $m_{AC} = \frac{4 - 1}{1 - 2} = \frac{3}{-1} = -3$

   Multiply the slopes of $AB$ and $AC$:

   $$m_{AB} \cdot m_{AC} = \frac{1}{3} \cdot (-3) = -1$$

   Since the product of their slopes is $-1$, sides $AB$ and $AC$ are perpendicular, proving that $\angle BAC$ is a right angle.