Analytical Geometry-G11-Maths-T4P2

Analytical Geometry

Analytical geometry uses algebra to study points, lines and shapes on the Cartesian plane. Coordinates, gradients, distances and equations can be used to prove geometrical relationships and calculate unknown lengths, angles and areas.

Distance between two points

For two points with coordinates:

A (x_{1}; y_{1}) and B (x_{2}; y_{2})

the distance between them is:

A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

The distance formula follows from the theorem of Pythagoras. Distance is always positive.

Example

Calculate the distance between the points:

A (- 2; 3) and B (4; - 5)

Substitute the coordinates into the distance formula.

A B = \sqrt{{(4 - (- 2))}^{2} + {(- 5 - 3)}^{2}}

A B = \sqrt{6^{2} + {(- 8)}^{2}}

A B = \sqrt{36 + 64}

A B = \sqrt{100}

A B = 10 units

Exam Tip

Keep negative values in brackets when substituting. Do not remove the square root until the sum of the squares has been simplified.

Midpoint of a line segment

The midpoint divides a line segment into two equal parts. For endpoints:

A (x_{1}; y_{1}) and B (x_{2}; y_{2})

the midpoint is:

M (\frac{x_{1} + x_{2}}{2}; \frac{y_{1} + y_{2}}{2})

Example

Determine the midpoint of the line segment joining:

P (- 3; 7) and Q (5; - 1)

M (\frac{- 3 + 5}{2}; \frac{7 + (- 1)}{2})

M (\frac{2}{2}; \frac{6}{2})

M (1; 3)

If the midpoint and one endpoint are known, substitute them into the midpoint formula to calculate the other endpoint.

Gradient

Gradient measures the steepness and direction of a straight line. For two distinct points on a line:

A (x_{1}; y_{1}) and B (x_{2}; y_{2})

the gradient is:

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

The change in the vertical coordinates is divided by the change in the horizontal coordinates.

A positive gradient means that the line rises from left to right. A negative gradient means that it falls from left to right. A horizontal line has a gradient of zero. The gradient of a vertical line is undefined.

Example

Calculate the gradient of the line through:

A (- 1; 2) and B (3; 10)

m_{A B} = \frac{10 - 2}{3 - (- 1)}

m_{A B} = \frac{8}{4}

m_{A B} = 2

Common Mistake

Use the coordinates in the same order in the numerator and denominator. If the second vertical coordinate is used first, the second horizontal coordinate must also be used first.

Parallel lines

Parallel lines have equal gradients.

m_{1} = m_{2}

Example

Determine whether the line through the first pair of points is parallel to the line through the second pair:

A (1; 2), B (4; 8), C (- 2; 1), D (0; 5)

m_{A B} = \frac{8 - 2}{4 - 1}

m_{A B} = \frac{6}{3} = 2

m_{C D} = \frac{5 - 1}{0 - (- 2)}

m_{C D} = \frac{4}{2} = 2

Therefore:

m_{A B} = m_{C D}

The two lines are parallel.

Perpendicular lines

The gradients of two non-vertical perpendicular lines are negative reciprocals. Their product is negative one.

m_{1} m_{2} = - 1

Equivalently:

m_{2} = - \frac{1}{m_{1}}

Example

A line has a gradient of:

m_{1} = \frac{3}{4}

Determine the gradient of a line perpendicular to it.

m_{2} = - \frac{1}{m_{1}}

m_{2} = - \frac{1}{\frac{3}{4}}

m_{2} = - \frac{4}{3}

Remember

A horizontal line and a vertical line are also perpendicular, although the gradient of the vertical line is undefined.

Collinear points

Points are collinear if they lie on the same straight line. This can be proved by showing that the gradients between different pairs of points are equal.

Example

Show that the following points are collinear:

A (- 1; 1), B (2; 7), C (4; 11)

m_{A B} = \frac{7 - 1}{2 - (- 1)}

m_{A B} = \frac{6}{3} = 2

m_{B C} = \frac{11 - 7}{4 - 2}

m_{B C} = \frac{4}{2} = 2

Since the gradients are equal and the line segments share the point B, the three points are collinear.

Equation of a straight line

The most commonly used equation of a straight line is:

y = m x + c

In this equation, the letter representing the gradient is the coefficient of the horizontal variable, while the constant represents the y-intercept.

Example

Determine the equation of the line with gradient negative two passing through the point:

P (3; 1)

Start with the general equation.

y = m x + c

Substitute the gradient.

y = - 2 x + c

Substitute the coordinates of the known point.

1 = - 2 (3) + c

1 = - 6 + c

c = 7

Therefore, the equation is:

y = - 2 x + 7

To determine the x-intercept, substitute zero for the vertical variable.

0 = - 2 x + 7

x = \frac{7}{2}

Therefore, the x-intercept is:

(\frac{7}{2}; 0)

The y-intercept is:

(0; 7)

Different forms of the equation of a line

Gradient-intercept form

This form shows the gradient and y-intercept directly.

y = m x + c

Point-gradient form

This form is useful when the gradient and one point are known.

y - y_{1} = m (x - x_{1})

Example

Find the equation of a line with gradient three passing through:

(- 2; 4)

y - 4 = 3 (x - (- 2))

y - 4 = 3 (x + 2)

y - 4 = 3 x + 6

y = 3 x + 10

General form

A straight-line equation may also be written as:

A x + B y + C = 0

For example:

2 x + 3 y - 12 = 0

Rearrange into gradient-intercept form to identify the gradient.

3 y = - 2 x + 12

y = - \frac{2}{3} x + 4

Therefore:

m = - \frac{2}{3}

Horizontal and vertical lines

A horizontal line has a constant vertical coordinate.

y = k

A vertical line has a constant horizontal coordinate.

x = k

Angle of inclination

The angle of inclination is measured anticlockwise from the positive horizontal axis to the line. Its value lies between zero degrees and one hundred and eighty degrees.

The gradient and angle of inclination are related by:

m = \tan θ

Therefore:

θ = \tan^{- 1} (m)

If the gradient is positive, the angle of inclination is acute. If the gradient is negative, the angle of inclination is obtuse. A calculator may give a negative reference angle for a negative gradient, so one hundred and eighty degrees must be added.

Example

Calculate the angle of inclination of a line with gradient two.

\tan θ = 2

θ = \tan^{- 1} (2)

θ = 63, 43^{°}

Example

Calculate the angle of inclination of a line with gradient negative one.

\tan θ = - 1

θ = - 45^{°}

The required angle must be between zero degrees and one hundred and eighty degrees.

θ = 180^{°} - 45^{°}

θ = 135^{°}

Calculating angles between lines

One method is to calculate the angle of inclination of each line and then find the positive difference.

α = | θ_{2} - θ_{1} |

The acute angle between two non-perpendicular lines may also be calculated using:

\tan α = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |

Example

Calculate the acute angle between lines with gradients one-half and two.

\tan α = | \frac{2 - \frac{1}{2}}{1 + (2) (\frac{1}{2})} |

\tan α = | \frac{\frac{3}{2}}{2} |

\tan α = \frac{3}{4}

α = \tan^{- 1} (\frac{3}{4})

α = 36, 87^{°}

The obtuse angle between the lines is supplementary to the acute angle.

180^{°} - 36, 87^{°} = 143, 13^{°}

Area problems involving coordinates

Coordinate area problems usually require the distance formula, the midpoint formula, equations of lines or perpendicular distances. Always draw a rough sketch and label the coordinates.

The area of a triangle is:

Area = \frac{1}{2} \times base \times perpendicular height

Example

Calculate the area of the triangle with vertices:

A (1; 2), B (7; 2), C (4; 6)

The line segment joining A and B is horizontal because its endpoints have equal vertical coordinates. Its length is:

A B = 7 - 1

A B = 6 units

The perpendicular height is the vertical distance from C to the line through A and B.

h = 6 - 2

h = 4 units

Area = \frac{1}{2} (6) (4)

Area = 12 square units

For a quadrilateral, divide the shape into two triangles, calculate each area and add them. Alternatively, identify a familiar shape such as a rectangle, parallelogram or trapezium and use the appropriate area formula.

Exam Tip

Do not assume that a line is a height merely because it appears vertical or perpendicular in a diagram. Prove perpendicularity using gradients when necessary. Give lengths in units, areas in square units and angles in degrees. Retain several decimal places during calculations and round only the final answer as instructed.

Study Guide For Grade

11

Mathematics