3.4 Coordinate Geometry

3.4 Coordinate Geometry

Coordinate geometry bridges algebra and geometry by representing geometric figures on the Cartesian coordinate plane. Master coordinate plane properties, formulas for distance, midpoint, and slope, linear equation formats, and parallel/perpendicular line relationships to excel on GRE Quantitative Reasoning.

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1. The Cartesian Coordinate Plane

The Cartesian coordinate plane is formed by two perpendicular number lines intersecting at the origin $(0,0)$:

• x-axis: Horizontal number line (positive to the right, negative to the left).
• y-axis: Vertical number line (positive upward, negative downward).

The Four Quadrants & Symmetry

       y-axis
         |
  Q II   |   Q I
 (-, +)  |  (+, +)
---------+--------- x-axis
  Q III  |   Q IV
 (-, -)  |  (+, -)
         |

• Symmetry Rules:
  • Reflection of $(x, y)$ across the x-axis yields $(x, -y)$.
  • Reflection of $(x, y)$ across the y-axis yields $(-x, y)$.
  • Reflection of $(x, y)$ across the origin yields $(-x, -y)$.

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2. Distance and Midpoint Formulas

The Distance Formula

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is derived directly from the Pythagorean theorem: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The Midpoint Formula

The midpoint $M$ of a line segment connecting $(x_1, y_1)$ and $(x_2, y_2)$ is the point whose coordinates are the arithmetic averages of the x- and y-coordinates: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Worked Example 3.4.1: Distance Between Two Points

Find the distance between points $A(-5, 2)$ and $B(7, -3)$.

Solution:

1. Apply distance formula: $d = \sqrt{(7 - (-5))^2 + (-3 - 2)^2}$.
2. Simplify differences: $7 - (-5) = 12$ and $-3 - 2 = -5$.
3. Square components: $12^2 = 144$ and $(-5)^2 = 25$.
4. Add and take square root: $d = \sqrt{144 + 25} = \sqrt{169} = 13$.

Worked Example 3.4.2: Finding a Missing Endpoint

Point $M(2, -3)$ is the midpoint of segment $AB$. If $A = (-4, 5)$, find the coordinates of endpoint $B(x_B, y_B)$.

Solution:

1. Set up midpoint equations: $\frac{-4 + x_B}{2} = 2 \quad \text{and} \quad \frac{5 + y_B}{2} = -3$
2. Solve for $x_B$: $-4 + x_B = 4 \implies x_B = 8$.
3. Solve for $y_B$: $5 + y_B = -6 \implies y_B = -11$. The coordinates of point $B$ are $(8, -11)$.

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3. Slope and Linear Equations

The slope ($m$) of a line measures its steepness and direction, defined as the ratio of vertical change ("rise") to horizontal change ("run"):

eq x_2)$$ ### Four Slope Orientations - **Positive Slope ($m > 0$):** Line rises from left to right. - **Negative Slope ($m < 0$):** Line falls from left to right. - **Zero Slope ($m = 0$):** Horizontal line ($y = c$). - **Undefined Slope:** Vertical line ($x = c$). ### Forms of Linear Equations 1. **Slope-Intercept Form:** $y = mx + b$ ($m = \text{slope}, b = \text{y-intercept}$). 2. **Standard Form:** $Ax + By = C$ ($A, B, C$ integers; slope $m = -\frac{A}{B}$ for $B eq 0$). 3. **Point-Slope Form:** $y - y_1 = m(x - x_1)$. --- ## 4. Parallel and Perpendicular Lines | Line Relationship | Condition on Slopes | Visual Example | | :--- | :--- | :--- | | **Parallel Lines** | Slopes are **equal**: $m_1 = m_2$ (different y-intercepts $b_1 eq b_2$) | $y = 2x + 3$ and $y = 2x - 7$ | | **Perpendicular Lines** | Slopes are **negative reciprocals**: $m_1 \cdot m_2 = -1 \implies m_2 = -\frac{1}{m_1}$ | $y = \frac{3}{4}x + 1$ and $y = -\frac{4}{3}x + 5$ | #### Worked Example 3.4.3: Perpendicular Line Equation Find the equation of the line passing through point $(3, -1)$ that is perpendicular to $y = -\frac{3}{4}x + 5$. **Solution:** 1. Given line slope is $m_1 = -\frac{3}{4}$. 2. Perpendicular slope is negative reciprocal: $m_2 = -\frac{1}{-3/4} = \frac{4}{3}$. 3. Use point-slope form with point $(3, -1)$: $$y - (-1) = \frac{4}{3}(x - 3) \implies y + 1 = \frac{4}{3}x - 4 \implies y = \frac{4}{3}x - 5$$ --- ## 5. Intercepts and Area Bounded by Lines - **X-intercept:** The point where the line crosses the x-axis ($y = 0$). - **Y-intercept:** The point where the line crosses the y-axis ($x = 0$). #### Worked Example 3.4.4: Area of Bounded Right Triangle Calculate the area of the triangle bounded by the line $3x + 4y = 24$, the x-axis, and the y-axis. **Solution:** 1. Find x-intercept (set $y = 0$): $3x + 4(0) = 24 \implies 3x = 24 \implies x = 8$. Intercept is $(8, 0)$. 2. Find y-intercept (set $x = 0$): $3(0) + 4y = 24 \implies 4y = 24 \implies y = 6$. Intercept is $(0, 6)$. 3. The right triangle has base $b = 8$ along the x-axis and height $h = 6$ along the y-axis. 4. Calculate Area: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 6 = 24 \text{ square units}$$ --- ## 6. GRE Coordinate Geometry Tips - **Diagrams are NOT to Scale:** GRE coordinate plane figures are drawn to demonstrate topology, not exact proportions. Never measure lengths or estimate angles visually on the screen—always calculate using formulas! - **Quick Sketching:** When presented with line intersection problems, make a 10-second scratchpad sketch to verify quadrant positions and slope signs.