# 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 )