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Geometry and Trigonometry - Coordinate Geometry in 2D and 3D

Grade 11IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The 2D Cartesian Plane and 3D Space: In 2D, points are defined by (x,y)(x, y) coordinates on a flat plane. In 3D, we introduce a third dimension, zz, resulting in coordinates (x,y,z)(x, y, z). Visually, while 2D is a flat surface like a piece of paper, 3D space can be imagined as a room where xx and yy represent the floor dimensions and zz represents the height from the floor.

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Gradient (Slope) of a Line: The gradient mm measures the steepness and direction of a line in 2D. A positive gradient slopes upwards from left to right, while a negative gradient slopes downwards. Visually, a larger absolute value of mm indicates a steeper line, and a gradient of zero represents a perfectly horizontal line.

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Parallel and Perpendicular Lines: Two lines are parallel if they have the same gradient (m1=m2m_1 = m_2), meaning they never intersect and maintain a constant distance apart. They are perpendicular if the product of their gradients is −1-1 (m1×m2=−1m_1 \times m_2 = -1), forming a 90∘90^{\circ} angle at their intersection point.

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Equations of a Straight Line: Lines in 2D can be expressed in gradient-intercept form y=mx+cy = mx + c, where cc is the yy-intercept (the point where the line crosses the vertical axis), or in general form ax+by+d=0ax + by + d = 0. Visually, the intercept cc shifts the line up or down the yy-axis without changing its tilt.

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Distance and Midpoint in 2D and 3D: The distance between two points is the length of the straight line segment connecting them. In 3D, this is the length of the space diagonal of a rectangular box formed by the coordinate differences. The midpoint is the exact center of this segment, calculated by averaging the coordinates of the endpoints.

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Coordinate Geometry of Circles: A circle in 2D is defined by its center (h,k)(h, k) and radius rr. Visually, it represents the set of all points that are exactly distance rr away from the center point. If the center is at the origin (0,0)(0,0), the equation simplifies significantly.

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Geometric Properties in 3D: Objects in 3D space, such as spheres or cuboids, are analyzed using coordinates. For example, the distance from a point (x,y,z)(x, y, z) to the origin is the length of the vector from (0,0,0)(0,0,0) to that point. Visually, the zz-axis is perpendicular to the xyxy-plane, creating a right-handed coordinate system.

📐Formulae

Distance between (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2): d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Midpoint in 2D: M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

Gradient of a line: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

Equation of a line (point-gradient form): y−y1=m(x−x1)y - y_1 = m(x - x_1)

Perpendicular gradient relationship: m1⋅m2=−1m_1 \cdot m_2 = -1

Distance between (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2): d=(x2−x1)2+(y2−y1)2+(z2−z1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Midpoint in 3D: M=(x1+x22,y1+y22,z1+z22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)

Equation of a circle: (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

💡Examples

Problem 1:

Find the equation of the perpendicular bisector of the line segment joining points A(2,−3)A(2, -3) and B(6,5)B(6, 5).

Solution:

  1. Find the midpoint of ABAB: M=(2+62,−3+52)=(4,1)M = \left( \frac{2+6}{2}, \frac{-3+5}{2} \right) = (4, 1). \ 2. Find the gradient of ABAB: mAB=5−(−3)6−2=84=2m_{AB} = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2. \ 3. Find the perpendicular gradient: m⊥=−1mAB=−12m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{2}. \ 4. Use the point-gradient form with M(4,1)M(4, 1) and m⊥=−12m_{\perp} = -\frac{1}{2}: y−1=−12(x−4)y - 1 = -\frac{1}{2}(x - 4). \ 5. Simplify: y−1=−12x+2  ⟹  y=−12x+3y - 1 = -\frac{1}{2}x + 2 \implies y = -\frac{1}{2}x + 3.

Explanation:

To find a perpendicular bisector, we first identify the point it must pass through (the midpoint) and its steepness (the negative reciprocal of the original line's gradient). Finally, we use these to construct the line equation.

Problem 2:

Calculate the distance between points P(1,−2,4)P(1, -2, 4) and Q(4,2,4)Q(4, 2, 4) in 3D space.

Solution:

  1. Identify the coordinates: (x1,y1,z1)=(1,−2,4)(x_1, y_1, z_1) = (1, -2, 4) and (x2,y2,z2)=(4,2,4)(x_2, y_2, z_2) = (4, 2, 4). \ 2. Apply the 3D distance formula: d=(4−1)2+(2−(−2))2+(4−4)2d = \sqrt{(4 - 1)^2 + (2 - (-2))^2 + (4 - 4)^2}. \ 3. Calculate internal values: d=(3)2+(4)2+(0)2d = \sqrt{(3)^2 + (4)^2 + (0)^2}. \ 4. Simplify: d=9+16+0=25d = \sqrt{9 + 16 + 0} = \sqrt{25}. \ 5. Result: d=5d = 5 units.

Explanation:

This problem uses the extension of the Pythagorean theorem into three dimensions. Since the zz-coordinates are the same, the distance calculation effectively reduces to a 2D problem on a plane parallel to the xyxy-plane.