Maths: Core 1 Chapter 6: Coordination Geometry and the Straight Line

coordinates in multiple dimemsions

- in 'n' dimensions and origin and 'n' coordinates are needed to identification

the mid point of a line segment

- midpoint of the line joining (x₁, y₁) and (x₂, y₂) is [(x₁+x₂)÷2, (y₁, y₂)÷2]

length of a line segment

- to calculate the length through horizontal and vertical vector components -if a segment AB has points A and B the length of line segment= A (x₁, y₁) B (x₂, y₂) AB²= (x₂-x₁)² + (y₂-y₁)² therefore, AB= √(x₂-x₁)² + (y₂-y₁)²

plane

- two dimensional -an origin and two coordinates needed (one for either of the directional axes) - two perpendicular axes

standard form

-Ax + By = C -A, B, C are integers (positive or negative whole numbers) -No fractions nor decimals in standard form. -Traditionally the "Ax" term is positive.

straight line equation forms

-a equation = a straight only if: y= mx+c (only y and x terms and a constant, no squares, reciprocals etc) -general form: ay+ bx+ c=0

line segment

-a finite part of a line -length can be calculated via numerical coordinates or algebraically

coordinate geometry of a plane

-horizontal axis (abscissa): x: Ox -vertical axis: y (ordinate): Oy -O= point of origin (where axes intersect), point 0 for both axes -point P's ordinate is it's +/- direction from the abscissa; P's abscissa is it's +/- directed distance from the ordinate

parabolas (equations on a curve)

-infinite number of points on a curve -tabulate for more accuracy and connect points through a smooth curve

gradient of the line

-measures how steep a line is -∆y÷ ∆x, m= (y₂-y₁)÷ (x₂-x₁) -can be positive, negative, zero or infinity

line

-one dimensional -on origin and one coordinate needed for identification -a straight line extending indefinitely in both directions

application

-two lines' intersection means that they have a point in common, to find this point, solve the equations simultaneously

straight line equation (given a point and it's gradient)

1. substitute the x and y values of the point given into the equation for m set to the actual value of m 2. solve for y point: (x₁, y₁) gradient of m y-y₁=m(x-x₁)

zero gradient

a line ‖ to the x-axis has no gradient

infinite gradient

a line ⊥ to the x-axis has a gradient of ∞

gradient as tan θ

m= tangent of the angle created between the line and the +ve abscissa - if tan θ is acute, m=+ve -in tan θ is obtuse, m=-ve

straight line equation (given two points)

point 1: (x₁, y₁) point 2 (x₂, y₂) method #1: (y-y₁)÷ (y₂-y₁)= (x-x₁)÷(x₂-x₁) Method #2: find the gradient (y₂-y₁)÷ (x₂-x₁), then substitute either point 1 or 2 into y= mx+c to solve for c, then put into y=mx+c

to test to see if a point lies on a curve

substitute the values of the coordinates into the equation, if the equation remains true, than the point exists on the curve

positive gradient

when both x and y increase m= +ve

negative gradient

when one axes increases as the other decreases m= -ve

parallel and perpendicular lines

‖ lines= same m ⊥ lines= inversely opposite m ie if two lines have the gradients m₁ and m₂ ⊥ if m₁×m₂= -1 ‖ if m₁=m₂