Straight Line Flashcards
Parallel lines…
have the same gradient
For perpendicular lines the following is true..
Example M qp and M qo
M qp = b M qo = - a
— And —
a b
• M qp X M qo = b -a
— X — = -1
a b
Gradient = m
M = y2 - y1
——–
x2 - x1
C = ?
Y intercept ( 0,c )
M = ?
Tan 0
Midpoint
Mid =
x1 + x2 y1 + y2
——- , ——-
2 2
How to find Median?
- Midpoint:
x1 + x2 y1 + y2 ------- , ------- 2 2 - Gradient:
y2 - y1 -------- x2 - x1 - Substitute into y - b= m( x - a )
How do you find the perpendicular bisector ?
- Midpoint:
x1 + x2 y1 + y2 ------- , ------- 2 2 - Gradient:
y2 - y1 -------- x2 - x1 - gradient of perpendicular bisector:
M qp X M qo = b -a
— X — = -1
a b
- substitute into y - b = m ( x - a)
Altitude of a triangle
- gradient:
y2 - y1 -------- x2 - x1 - perpendicular gradient:
M qp X M qo = b -a
— X — = -1
a b
- substitute into y - b = m ( x - a )
Median of a triangle
- midpoint:
x1 + x2 y1 + y2 ------- , ------- 2 2 - gradient:
y2 - y1 -------- x2 - x1 - substitute into y - b = m ( x - a )
Distance between 2 points
D = (x2 - x1)^2 + ( y2 - y1) ^2
Form for finding line equation
y - b = m (x - a )
( a, b) = point on the line
Concurrent lines
- Simultaneous equations
- Find y
Sub in y into equation - Sub in again to find x
- To find the lines pass through ( a, b ) so they are concurrent
Intersecting lines
- Median
- midpoint
- gradient
- equation
- simplify
- Altitude
- gradient
- perpendicular gradient
- equation
- simplify
- Simultaneous equation
- sub in to find points ( x, y )
