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Analytical geometry

Introduction

Analytical geometry, also called co-ordinate geometry and earlier referred to as Cartesian geometry, is the study of geometry using the principles of algebra, and the Cartesian co-ordinate system. It is concerned with defining geometrical shapes in a numerical way, and extracting numerical information from that representation. Some consider that the introduction of analytic geometry was the beginning of modern mathematics.

Distance between two points

One of the simplest things that can be done with analytical geometry is to calculate the distance between two points. Distance is a number that describes how far apart two point are. For example, point P has co-ordinates ( 2 , 1 ) and point Q has co-ordinates ( - 2 , - 2 ) . How far apart are points P and Q ? In the figure, this means how long is the dashed line?

In the figure, it can be seen that the length of the line P R is 3 units and the length of the line Q R is four units. However, the P Q R , has a right angle at R . Therefore, the length of the side P Q can be obtained by using the Theorem of Pythagoras:

P Q 2 = P R 2 + Q R 2 P Q 2 = 3 2 + 4 2 P Q = 3 2 + 4 2 = 5

The length of P Q is the distance between the points P and Q .

In order to generalise the idea, assume A is any point with co-ordinates ( x 1 ; y 1 ) and B is any other point with co-ordinates ( x 2 ; y 2 ) .

The formula for calculating the distance between two points is derived as follows. The distance between the points A and B is the length of the line A B . According to the Theorem of Pythagoras, the length of A B is given by:

A B = A C 2 + B C 2

However,

B C = y 2 - y 1 A C = x 2 - x 1

Therefore,

A B = A C 2 + B C 2 = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2

Therefore, for any two points, ( x 1 ; y 1 ) and ( x 2 ; y 2 ) , the formula is:

Distance= ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2

Using the formula, distance between the points P and Q with co-ordinates (2;1) and (-2;-2) is then found as follows. Let the co-ordinates of point P be ( x 1 ; y 1 ) and the co-ordinates of point Q be ( x 2 ; y 2 ) . Then the distance is:

Distance = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 2 - ( - 2 ) ) 2 + ( 1 - ( - 2 ) ) 2 = ( 2 + 2 ) 2 + ( 1 + 2 ) 2 = 16 + 9 = 25 = 5

The following video provides a summary of the distance formula.

Khan academy video on distance formula

Calculation of the gradient of a line

The gradient of a line describes how steep the line is. In the figure, line P T is the steepest. Line P S is less steep than P T but is steeper than P R , and line P R is steeper than P Q .

The gradient of a line is defined as the ratio of the vertical distance to the horizontal distance. This can be understood by looking at the line as the hypotenuse of a right-angled triangle. Then the gradient is the ratio of the length of the vertical side of the triangle to the horizontal side of the triangle. Consider a line between a point A with co-ordinates ( x 1 ; y 1 ) and a point B with co-ordinates ( x 2 ; y 2 ) .

Gradient = y 2 - y 1 x 2 - x 1

We can use the gradient of a line to determine if two lines are parallel or perpendicular. If the lines are parallel ( [link] a) then they will have the same gradient, i.e. m AB = m CD . If the lines are perpendicular ( [link] b) than we have: - 1 m AB = m CD

For example the gradient of the line between the points P and Q , with co-ordinates (2;1) and (-2;-2) ( [link] ) is:

Gradient = y 2 - y 1 x 2 - x 1 = - 2 - 1 - 2 - 2 = - 3 - 4 = 3 4

The following video provides a summary of the gradient of a line.

Gradient of a line

Midpoint of a line

Sometimes, knowing the co-ordinates of the middle point or midpoint of a line is useful. For example, what is the midpoint of the line between point P with co-ordinates ( 2 ; 1 ) and point Q with co-ordinates ( - 2 ; - 2 ) .

The co-ordinates of the midpoint of any line between any two points A and B with co-ordinates ( x 1 ; y 1 ) and ( x 2 ; y 2 ) , is generally calculated as follows. Let the midpoint of A B be at point S with co-ordinates ( X ; Y ) . The aim is to calculate X and Y in terms of ( x 1 ; y 1 ) and ( x 2 ; y 2 ) .

X = x 1 + x 2 2 Y = y 1 + y 2 2 S x 1 + x 2 2 ; y 1 + y 2 2

Then the co-ordinates of the midpoint ( S ) of the line between point P with co-ordinates ( 2 ; 1 ) and point Q with co-ordinates ( - 2 ; - 2 ) is:

X = x 1 + x 2 2 = - 2 + 2 2 = 0 Y = y 1 + y 2 2 = - 2 + 1 2 = - 1 2 S is at ( 0 ; - 1 2 )

It can be confirmed that the distance from each end point to the midpoint is equal. The co-ordinate of the midpoint S is ( 0 ; - 0 , 5 ) .

P S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - 2 ) 2 + ( - 0 . 5 - 1 ) 2 = ( - 2 ) 2 + ( - 1 . 5 ) 2 = 4 + 2 . 25 = 6 . 25

and

Q S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - ( - 2 ) ) 2 + ( - 0 . 5 - ( - 2 ) ) 2 = ( 0 + 2 ) ) 2 + ( - 0 . 5 + 2 ) ) 2 = ( 2 ) ) 2 + ( - 1 . 5 ) ) 2 = 4 + 2 . 25 = 6 . 25

It can be seen that P S = Q S as expected.

The following video provides a summary of the midpoint of a line.

Khan academy video on midpoint of a line

Co-ordinate geometry

  1. In the diagram given the vertices of a quadrilateral are F(2;0), G(1;5), H(3;7) and I(7;2).
    1. What are the lengths of the opposite sides of FGHI?
    2. Are the opposite sides of FGHI parallel?
    3. Do the diagonals of FGHI bisect each other?
    4. Can you state what type of quadrilateral FGHI is? Give reasons for your answer.
  2. A quadrialteral ABCD with vertices A(3;2), B(1;7), C(4;5) and D(1;3) is given.
    1. Draw the quadrilateral.
    2. Find the lengths of the sides of the quadrilateral.
  3. ABCD is a quadrilateral with verticies A(0;3), B(4;3), C(5;-1) and D(-1;-1).
    1. Show that:
      1. AD = BC
      2. AB DC
    2. What name would you give to ABCD?
    3. Show that the diagonals AC and BD do not bisect each other.
  4. P, Q, R and S are the points (-2;0), (2;3), (5;3), (-3;-3) respectively.
    1. Show that:
      1. SR = 2PQ
      2. SR PQ
    2. Calculate:
      1. PS
      2. QR
    3. What kind of a quadrilateral is PQRS? Give reasons for your answers.
  5. EFGH is a parallelogram with verticies E(-1;2), F(-2;-1) and G(2;0). Find the co-ordinates of H by using the fact that the diagonals of a parallelogram bisect each other.

Questions & Answers

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Graphene has a hexagonal structure
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Source:  OpenStax, Siyavula textbooks: grade 10 maths [ncs]. OpenStax CNX. Aug 05, 2011 Download for free at http://cnx.org/content/col11239/1.2
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