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
