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Geometry and Trigonometry - Coordinate geometry

Grade 12IB_AI

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

🔑Concepts

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The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is the length of the hypotenuse of a right-angled triangle formed by the horizontal and vertical differences.

A right-angled triangle illustrating the distance formula between two points A and B.
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The midpoint MM of a line segment is the average of the xx-coordinates and the average of the yy-coordinates of the endpoints.

A line segment with its midpoint M clearly marked.
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Parallel lines have identical gradients (m1=m2m_1 = m_2), while perpendicular lines have gradients that are negative reciprocals (m1×m2=−1m_1 \times m_2 = -1).

Two lines intersecting at a 90-degree angle, demonstrating perpendicular gradients.
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In 3D coordinate geometry, points are defined by (x,y,z)(x, y, z). Formulas for distance and midpoint extend logically to include the zz component.

📐Formulae

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

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

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

y−y1=m(x−x1)y - y_1 = m(x - x_1)

m1×m2=−1m_1 \times m_2 = -1

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}

M3D=(x1+x22,y1+y22,z1+z22)M_{3D} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)

💡Examples

Problem 1:

Find the equation of the perpendicular bisector of the line segment joining A(2,4)A(2, 4) and B(6,10)B(6, 10).

Solution:

  1. Find the midpoint of ABAB: M=(2+62,4+102)=(4,7)M = \left( \frac{2 + 6}{2}, \frac{4 + 10}{2} \right) = (4, 7) 2. Find the gradient of ABAB: mAB=10−46−2=64=1.5m_{AB} = \frac{10 - 4}{6 - 2} = \frac{6}{4} = 1.5 3. Find the perpendicular gradient: m⊥=−11.5=−23m_{\perp} = -\frac{1}{1.5} = -\frac{2}{3} 4. Use the point-gradient formula with M(4,7)M(4, 7): y−7=−23(x−4)y - 7 = -\frac{2}{3}(x - 4) y=−23x+83+7⇒y=−23x+293y = -\frac{2}{3}x + \frac{8}{3} + 7 \Rightarrow y = -\frac{2}{3}x + \frac{29}{3}

Explanation:

A perpendicular bisector must pass through the midpoint of the segment and have a gradient that is the negative reciprocal of the original line's gradient.

Problem 2:

Calculate the distance between the points P(1,−2,5)P(1, -2, 5) and Q(4,2,5)Q(4, 2, 5) in 3D space.

Solution:

Using the 3D distance formula: d=(4−1)2+(2−(−2))2+(5−5)2d = \sqrt{(4 - 1)^2 + (2 - (-2))^2 + (5 - 5)^2} d=32+42+02d = \sqrt{3^2 + 4^2 + 0^2} d=9+16+0=25=5d = \sqrt{9 + 16 + 0} = \sqrt{25} = 5

Explanation:

The distance is found by applying the Pythagorean theorem to the differences in xx, yy, and zz coordinates. Since the zz-coordinates are the same, the distance is effectively calculated in the xyxy-plane.

Problem 3:

A Voronoi diagram has two sites at A(0,2)A(0, 2) and B(4,2)B(4, 2). Determine the equation of the edge that separates these two sites.

Solution:

The edge between two sites in a Voronoi diagram is the perpendicular bisector of the segment connecting them.

  1. Midpoint of ABAB: M=(0+42,2+22)=(2,2)M = \left( \frac{0+4}{2}, \frac{2+2}{2} \right) = (2, 2)
  2. Gradient of ABAB: m=2−24−0=0m = \frac{2 - 2}{4 - 0} = 0
  3. Since the line ABAB is horizontal (m=0m=0), the perpendicular bisector must be a vertical line passing through the xx-coordinate of the midpoint. Equation: x=2x = 2

Explanation:

In a Voronoi diagram, the boundary between two sites is equidistant from both. For sites (0,2)(0, 2) and (4,2)(4, 2), the vertical line x=2x=2 splits the plane such that any point on the left is closer to AA and any point on the right is closer to BB.

Problem 4:

Find the equation of the line passing through point C(3,−1)C(3, -1) that is parallel to the line y=2x+5y = 2x + 5.

Two parallel lines on a coordinate plane, one passing through point C.

Solution:

  1. Identify the gradient of the given line: m=2m = 2.
  2. Since the lines are parallel, the gradient of the new line is also m=2m = 2.
  3. Use the point-gradient form y−y1=m(x−x1)y - y_1 = m(x - x_1) with (x1,y1)=(3,−1)(x_1, y_1) = (3, -1): y−(−1)=2(x−3)y - (-1) = 2(x - 3) y+1=2x−6y + 1 = 2x - 6 y=2x−7y = 2x - 7

Explanation:

Parallel lines must have the same slope to ensure they never meet. We use the given point and this slope to find the unique intercept.

Problem 5:

A triangle has vertices at L(1,1,2)L(1, 1, 2), M(1,5,2)M(1, 5, 2), and N(4,1,2)N(4, 1, 2). Calculate the perimeter of triangle LMNLMN.

A right-angled triangle LMN plotted on a coordinate plane.

Solution:

LM=(1−1)2+(5−1)2+(2−2)2=02+42+02=4LM = \sqrt{(1-1)^2 + (5-1)^2 + (2-2)^2} = \sqrt{0^2 + 4^2 + 0^2} = 4 LN=(4−1)2+(1−1)2+(2−2)2=32+02+02=3LN = \sqrt{(4-1)^2 + (1-1)^2 + (2-2)^2} = \sqrt{3^2 + 0^2 + 0^2} = 3 MN=(4−1)2+(1−5)2+(2−2)2=32+(−4)2+02=9+16=5MN = \sqrt{(4-1)^2 + (1-5)^2 + (2-2)^2} = \sqrt{3^2 + (-4)^2 + 0^2} = \sqrt{9+16} = 5 Perimeter=4+3+5=12 units\text{Perimeter} = 4 + 3 + 5 = 12 \text{ units}

Explanation:

Since all points have the same zz-coordinate (z=2z=2), this triangle lies on a horizontal plane. We calculate the lengths of LMLM, MNMN, and NLNL using the 3D distance formula and sum them.