Revision (Grade 11 NSC Matric Mathematics): Revision Notes

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What is analytical geometry?

Analytical geometry, also called coordinate geometry or Cartesian geometry, is the study of geometric properties and relationships between points, lines and angles in the Cartesian plane. In analytical geometry, we define geometric shapes using a coordinate system and algebraic principles. This branch of mathematics allows us to solve geometric problems using equations and formulas.

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Analytical geometry bridges the gap between algebra and geometry, making it possible to solve complex geometric problems using coordinate systems and algebraic methods. This approach is essential in fields like engineering, physics, and computer graphics.

Distance formula and theorem of Pythagoras

The distance formula is based on the theorem of Pythagoras and helps us find the length between any two points on a coordinate plane.

Figure

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Critical Foundation: The distance formula is essentially the Pythagorean theorem applied to coordinates. Understanding this connection helps you remember and apply the formula correctly.

Theorem of Pythagoras: AB2=AC2+BC2AB^2 = AC^2 + BC^2

Distance formula: For two points with coordinates (x1;y1)(x_1; y_1) and (x2;y2)(x_2; y_2), the distance between them is:

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

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Remember that (x1−x2)2=(x2−x1)2(x_1 - x_2)^2 = (x_2 - x_1)^2, so the order of subtraction doesn't matter when we square the differences. This makes the formula flexible and reduces calculation errors.

Gradient and slope

Gradient (also called slope) measures how steep a line is. It tells us the rate of change between two points on a line.

Gradient formula: For a line passing through points (x1;y1)(x_1; y_1) and (x2;y2)(x_2; y_2):

mAB=y2−y1x2−x1 or mAB=y1−y2x1−x2m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \text{ or } m_{AB} = \frac{y_1 - y_2}{x_1 - x_2}

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Common Mistake to Avoid: Be consistent with your formula choice. Remember: m≠y1−y2x2−x1m \neq \frac{y_1 - y_2}{x_2 - x_1}

Always keep the order of subtraction the same for both numerator and denominator.

Types of lines and their gradients

Different types of lines have specific gradient values that help us identify and work with them.

Figure

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Line Types and Their Gradients:

  • Horizontal lines: These lines run left to right and have gradient m=0m = 0
  • Vertical lines: These lines run up and down and have gradient mm that is undefined
  • Parallel lines: These lines never meet and have equal gradients: m1=m2m_1 = m_2
  • Perpendicular lines: These lines meet at right angles and their gradients multiply to give −1-1: m1×m2=−1m_1 \times m_2 = -1

Mid-point of a line segment

The mid-point is the point that lies exactly halfway between two given points on a line segment.

Figure

Mid-point formula: For two points (x1;y1)(x_1; y_1) and (x2;y2)(x_2; y_2), the mid-point M(x;y)M(x; y) is:

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

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This formula simply averages the x-coordinates and y-coordinates of the two points. Think of it as finding the "middle" by taking the average of each coordinate separately.

Points on a straight line (collinear points)

When three or more points lie on the same straight line, we call them collinear points. For any point (x;y)(x; y) on a straight line passing through points (x1;y1)(x_1; y_1) and (x2;y2)(x_2; y_2):

Figure

y−y1x−x1=y2−y1x2−x1\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}

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This equation tells us that the gradient between any two points on the same straight line is constant. This is a fundamental property that helps us determine if points are collinear.

Quadrilaterals and their properties

A quadrilateral is a closed shape with four straight line segments. Different types of quadrilaterals have specific properties.

Parallelogram

Figure

A parallelogram is a quadrilateral where both pairs of opposite sides are parallel.

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Parallelogram Properties:

  • Both pairs of opposite sides are equal in length
  • Both pairs of opposite angles are equal
  • The diagonals bisect each other

Rectangle

Figure

A rectangle is a parallelogram with all four angles equal to 90°90°.

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Rectangle Properties:

  • Both pairs of opposite sides are equal and parallel
  • The diagonals bisect each other
  • The diagonals are equal in length

Rhombus

Figure

A rhombus is a parallelogram with all four sides equal in length.

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Rhombus Properties:

  • Both pairs of opposite sides are equal and parallel
  • The diagonals bisect each other at 90°90°
  • The diagonals of a rhombus bisect both pairs of opposite angles

Square

A square is a rhombus with all four interior angles equal to 90°90°.

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Square Properties:

  • Both pairs of opposite sides are equal and parallel
  • The diagonals bisect each other at 90°90°
  • The diagonals are equal in length
  • The diagonals bisect both pairs of interior opposite angles (all angles are 45°45°)

Trapezium

Figure

A trapezium is a quadrilateral with one pair of opposite sides parallel.

Kite

A kite is a quadrilateral with two pairs of adjacent sides equal.

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Kite Properties:

  • One pair of opposite angles are equal (angles between unequal sides)
  • The diagonal between equal sides bisects the other diagonal
  • The diagonal between equal sides bisects the interior angles
  • The diagonals intersect at 90°90°

Worked example 1: Basic calculations

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Worked Example: Distance, Mid-point and Perpendicular Lines

Question: Given points P(−5;−4)P(-5; -4) and Q(0;6)Q(0; 6):

  1. Find the length of line segment PQPQ
  2. Find the mid-point T(x;y)T(x; y) of line segment PQPQ
  3. Show that the line through R(1;−34)R(1; -\frac{3}{4}) and T(x;y)T(x; y) is perpendicular to line PQPQ

Solution:

Step 1: Assign coordinates

  • Let P(x1;y1)=P(−5;−4)P(x_1; y_1) = P(-5; -4) and Q(x2;y2)=Q(0;6)Q(x_2; y_2) = Q(0; 6)
  • So: x1=−5x_1 = -5, y1=−4y_1 = -4, x2=0x_2 = 0, y2=6y_2 = 6

Step 2: Apply the distance formula

  • PQ=(x2−x1)2+(y2−y1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  • =(0−(−5))2+(6−(−4))2= \sqrt{(0 - (-5))^2 + (6 - (-4))^2}
  • =25+100=125=55= \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5}

The length of line segment PQPQ is 555\sqrt{5} units.

Step 3: Apply the mid-point formula

  • T(x;y)=(x1+x22;y1+y22)T(x; y) = \left(\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}\right)

For x-coordinate: x=−5+02=−52x = \frac{-5 + 0}{2} = -\frac{5}{2}

For y-coordinate: y=−4+62=1y = \frac{-4 + 6}{2} = 1

The mid-point is T(−52;1)T(-\frac{5}{2}; 1).

Step 4: Calculate gradients and check perpendicularity

  • Gradient of PQPQ: mPQ=6−(−4)0−(−5)=105=2m_{PQ} = \frac{6 - (-4)}{0 - (-5)} = \frac{10}{5} = 2

  • Gradient of RTRT: mRT=34−11−(−52)=−7472=−12m_{RT} = \frac{\frac{3}{4} - 1}{1 - (-\frac{5}{2})} = \frac{-\frac{7}{4}}{\frac{7}{2}} = -\frac{1}{2}

  • Product of gradients: mRT×mPQ=−12×2=−1m_{RT} \times m_{PQ} = -\frac{1}{2} \times 2 = -1

Since the product equals −1-1, line PQPQ is perpendicular to line RTRT.

Worked example 2: Quadrilaterals

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Worked Example: Finding the Fourth Vertex of a Parallelogram

Question: Points A(−1;0)A(-1; 0), B(0;3)B(0; 3), C(8;11)C(8; 11) and D(x;y)D(x; y) are points on the Cartesian plane. Find D(x;y)D(x; y) if ABCDABCD is a parallelogram.

Figure

Solution:

Step 1: Use the property that diagonals of a parallelogram bisect each other

In a parallelogram, the mid-point of diagonal ACAC equals the mid-point of diagonal BDBD.

Step 2: Find the mid-point of ACAC

Let A(−1;0)A(-1; 0) and C(8;11)C(8; 11)

  • MAC=(−1+82;0+112)=(72;112)M_{AC} = \left(\frac{-1 + 8}{2}; \frac{0 + 11}{2}\right) = \left(\frac{7}{2}; \frac{11}{2}\right)

Step 3: Use the fact that this equals the mid-point of BDBD

Since B(0;3)B(0; 3) and D(x;y)D(x; y):

  • MBD=(0+x2;3+y2)M_{BD} = \left(\frac{0 + x}{2}; \frac{3 + y}{2}\right)

Step 4: Set the mid-points equal and solve

  • (72;112)=(0+x2;3+y2)\left(\frac{7}{2}; \frac{11}{2}\right) = \left(\frac{0 + x}{2}; \frac{3 + y}{2}\right)

For x-coordinate: 72=x2\frac{7}{2} = \frac{x}{2}, so x=7x = 7

For y-coordinate: 112=3+y2\frac{11}{2} = \frac{3 + y}{2}, so 11=3+y11 = 3 + y, therefore y=8y = 8

The coordinates of DD are (7;8)(7; 8).

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Key Points to Remember:

  • The distance formula uses Pythagoras: (x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
  • Perpendicular lines have gradients that multiply to give −1-1
  • Parallel lines have equal gradients
  • The mid-point formula averages the coordinates: (x1+x22;y1+y22)\left(\frac{x_1+x_2}{2}; \frac{y_1+y_2}{2}\right)
  • In parallelograms, diagonals bisect each other, making their mid-points equal

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