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Fun sums

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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)Equation?

m = 11–7/2–1 = 4/1 = 4

y = 4x + ?

(x,y)

(a,b)

m = y – b

x – a

m(x – a) = y – b

y – b = m(x – a)

Equation in ActionLine 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

y – b = m(x – a)

where m → gradient

(a , b) → any point on line

Ex 1 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)

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

= 2/3

Key Question

Find the equation of the line connecting

A (2 , 6) and B (5 , 8)

m = 8 – 6

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

5 – 2 = 2/3

Key Question

Find the equation of the line connecting

A (2 , 6) and B (5 , 8)

m = 8 – 6

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

5 – 2 = 2/3

Key Question 2 C is the mid point of AB and CD is perpendicular to AB

Find the equation of CD.

mAB = 10 – 5

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

5 – 2 = 5/3 A (2 , 5)

B(5 , 10)

D

C(3½ , 7½)

m = -3/5