Download Equation of a Straight Line

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Equation of a Straight Line
For any line!
What is to be learned?
• How to find equation of straight line given
any two points
A (0 , 7 )
B (2 , 11)
Equation?
m = 11–7/2–0 = 4/2 = 2
y = 2x + 7 (using y = mx +c ) 
A (1 , 7 )
B (2 , 11)
m = 11–7/2–1 = 4/1 = 4
y = 4x + ? 
Equation?
(x,y)
(a,b)
m= y–b
x–a
m(x – a) = y – b
y – b = m(x – a)
Equation in Action
Line has gradient 5 and contains
point (-4 , 3)
m=5
(a , b) = (-4 , 3)
y – b = m(x – a)
y – 3 = 5(x + 4)
y – 3 = 5x + 20
y = 5x + 20 + 3
y = 5x + 23
Equation of any Straight Line
use
Ex 1
y – b = m(x – a)
where m → gradient
(a , b) → any point on line
Line has gradient 4 and contains
point (2 , 6)
m=4
(a , b) = (2 , 6)
y – 6 = 4(x – 2)
y – 6 = 4x – 8
y = 4x – 2
Ex 2
Line contains points (3 , -4)
and (6, -2). Equation?
Need gradient and point!
m = -2 – (-4)
= 2/3
6–3
using m = 2/3 and (a , b) = (3 , -4)
y – (-4) = 2/3(x – 3)
X3→
3y + 12 = 2(x – 3)
3y + 12 = 2x – 6
3y = 2x – 18
Key Question
Find the equation of the line connecting
A (2 , 6) and B (5 , 8)
m = 8 – 6 = 2/3
5–2
y – b = m(x – a) with m = 2/3 , (a , b) = (2 , 6)
y – 6 = 2/3 (x – 2)
3y – 18 = 2(x – 2)
3y – 18 = 2x – 4
3y = 2x + 14
Key Question
Find the equation of the line connecting
A (2 , 6) and B (5 , 8)
m = 8 – 6 = 2/3
5–2
y – b = m(x – a) with m = 2/3 , (a , b) = (5 , 8)
y – 8 = 2/3 (x – 5)
3y – 24 = 2(x – 5)
3y – 24 = 2x – 10
3y = 2x + 14
Key Question 2
C is the mid point of AB and CD is perpendicular to AB
Find the equation of CD.
C
(3½ , 7½)B(5 , 10)
mAB = 10 – 5 = 5/
A (2 , 5)
3
5–2
mCD = -3/5 (m1m2 = -1)
C = (3½ , 7½) (mid pt)
m = -3/5 , (a , b) = (3½ , 7½)
y – b = m(x – a)
y – 7½ = -3/5(x – 3½)
5y – 37½ = -3(x – 3½)
5y – 37½ = -3x + 10½
5y = -3x + 48
m = -3/5
D