5.4 Coordinate Geometry

Coordinate Plane Basics

Coordinate geometry blends algebra and geometry, allowing us to describe geometric shapes using algebraic equations on a grid. On the SHSAT, you will be expected to calculate slopes, find distances and midpoints between points, and determine the areas of shapes plotted on the coordinate plane.

1. The Coordinate Plane and Plotting Points

The coordinate plane is formed by the intersection of a horizontal number line (the x-axis) and a vertical number line (the y-axis). The point where they intersect is the origin, represented by the coordinates (0, 0).

The axes divide the plane into four quadrants, numbered counterclockwise using Roman numerals:

• Quadrant I: Top-right (positive x, positive y)
• Quadrant II: Top-left (negative x, positive y)
• Quadrant III: Bottom-left (negative x, negative y)
• Quadrant IV: Bottom-right (positive x, negative y)

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2. Slope of a Line

Slope (m) measures the steepness and direction of a line. It is defined as the ratio of the vertical change (rise) to the horizontal change (run) between any two points (x₁, y₁) and (x₂, y₂) on the line:

$$m = \frac{\text{Change in } y}{\text{Change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$$

Types of Slopes

• Positive Slope: The line rises from left to right (as x increases, y increases).
• Negative Slope: The line falls from left to right (as x increases, y decreases).
• Zero Slope: The line is horizontal (the y-values are equal, so y₂ - y₁ = 0).
• Undefined Slope: The line is vertical (the x-values are equal, so x₂ - x₁ = 0, leading to division by zero).

Parallel and Perpendicular Lines

The relationship between the slopes of two lines determines whether they are parallel or perpendicular:

• Parallel Lines: Lines that never intersect. Parallel lines have equal slopes (m_1 = m_2).
• Perpendicular Lines: Lines that intersect at a 90° angle. Perpendicular lines have slopes that are negative reciprocals (m_1 * m_2 = -1). For example, if a line has a slope of 2/3, a line perpendicular to it will have a slope of -3/2.

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3. The Distance Formula

The distance between any two points (x₁, y₁) and (x₂, y₂) on a coordinate plane is the length of the straight line segment connecting them. This is calculated using the Distance Formula:

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

The Pythagorean Connection

The distance formula is directly derived from the Pythagorean Theorem. If you draw horizontal and vertical lines from your two points, you form a right triangle. The horizontal distance is the change in x (Δx = x₂ - x₁), and the vertical distance is the change in y (Δy = y₂ - y₁). The straight-line distance is the hypotenuse:

$$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$

Worked Example: Find the distance between the points A(-3, 4) and B(9, -1).

Solution:

• Identify coordinates: x₁ = -3, y₁ = 4 and x₂ = 9, y₂ = -1.
• Find the differences: Δx = 9 - (-3) = 12 Δy = -1 - 4 = -5
• Plug into the formula: d = √((12)² + (-5)²) d = √(144 + 25) = √169 = 13. The distance between points A and B is 13 units (recalling the 5-12-13 Pythagorean triple).

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4. The Midpoint Formula

The midpoint of a line segment is the point that is exactly halfway between the two endpoints (x₁, y₁) and (x₂, y₂). The coordinates of the midpoint (M) are found by averaging the x-coordinates and averaging the y-coordinates of the endpoints:

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

Worked Example: Segment AB has endpoints A(2, -3) and B(8, 7). Find the midpoint M. Solution:

• Average the x-coordinates: (2 + 8) / 2 = 5
• Average the y-coordinates: (-3 + 7) / 2 = 2
• The midpoint is M(5, 2).

The Endpoint Trap

A classic SHSAT trick is to give you one endpoint and the midpoint, and ask you to find the other endpoint. Problem: The midpoint of segment CD is M(3, 1). If endpoint C has coordinates (-1, 5), what are the coordinates of endpoint D(x₂, y₂)?

Strategy: Write down the average equations and solve for the unknown coordinates:

• Midpoint x: (-1 + x_2) / 2 = 3 => -1 + x_2 = 6 => x_2 = 7
• Midpoint y: (5 + y_2) / 2 = 1 => 5 + y_2 = 2 => y_2 = -3 Endpoint D is at (7, -3).

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5. Finding Area on the Grid: Bounding Box Method

When asked to find the area of an irregular polygon (like a slanted triangle or trapezoid) on a coordinate plane, counting grid squares is inaccurate, and calculating lengths using the distance formula can be slow and error-prone. The Bounding Box Method is the most reliable strategy.

How to Use the Bounding Box Method:

1. Draw a rectangle (a bounding box) enclosing the shape. The sides of the rectangle must run along the horizontal and vertical grid lines.
2. The corners of the bounding box will align with the outermost coordinates of the shape.
3. Calculate the area of this large bounding box (Area_box = length * width).
4. The spaces outside your shape but inside the bounding box will form simple right triangles and/or rectangles. Calculate the area of each of these outer shapes.
5. Subtract the areas of these outer shapes from the area of the bounding box to find the area of the inner shape.