2 - 1 / 1 - 4 = - 1/3

2 = -1/3(1) + b

6/3 = -1/3 + b

7/3 = b

y = -1/3x + b

3y = -x + 21/3, 3y = -x + 7

x + 3y = 7 (standard form)

The typical equation of a line is: y = mx + y₀ → where m: slope and where b: y₀-intercept

How to get the equation of the line that passes through A (1; 2)   B (4 ; 1) ?

To calculate m

m = (yB - yA) / (xB - xA) = (1 - 2) / (4 - 1) = - 1/3

The equation of the line (AB) becomes: y = - (1/3).x + y₀

To calculate b

The line passes through A, so the coordinates of this point must verify the equation of the line.

y = - (1/3).x + y₀

y₀ = y + (1/3).x → you substitute x and y by the coordinates of the point A (1 ; 2)

y₀ = 2 + [(1/3) * 1]

y₀ = 2 + (1/3)

y₀ = 7/3

The equation of the line (AB) is: y = - (1/3).x + (7/3) → to go further

y = - (1/3).x + (7/3)

y = (- x + 7)/3

3y = - x + 7

x + 3y = 7

Points: (1, 2) and (4, 1)

Slope: -1/3

Straight line equation: y = -1/3x +7/3

:-

m = - 1/3

y - 2 = (-1/3) ( x - 1 )

y = (-1/3) x + 7/3

(1, 2) (4, 1)

ax + by + c = 0 … (1)equation (1) passing (1,2) and (4,1)ax + by + c = 0a (1) + b (2) + c = 0a + 2b + c = 0 …. (2)ax + by + c = 0a (4) + b (1) + c = 04a + b + c = 0 …. (3)Solving equation (2) and (3)a + 2b + c = 04a + b + c = 0(-)    (-)  (-)_____________-3a + b = 0-3a = - b3a = ba/b = 1/3a = 1 and b = 3from equation (2)a + 2b + c = 01 + 2(3) + c = 01 + 6 + c = 0c = -7(1)x + (3)y – 7 = 0 … (1)x + 3y – 7 = 0

(1,2)(4,1)

f(x) = mx + b

m = (1-2)/(4-1) = -1/3

2  = (-1/3)(1) + b

b = 7/3

f(x) = (-1/3)x +7/3

f(6) = (-1/3)(6) +7/3

f(6) = 1/3

M= 1/3

F (x)= mx +b

F (4)= 1/3 (4) + b

F(4) - 4/3= 8/3 

F(x)=1/3x + 8/3 

Is this the correct answer?? 

x*m + c = y 

4m + c = 1 

m + c = 2 

m = -1/3, c = 7/3 

y = (7 – x)/ 

y = mx + b

m = (y2-y1)/(x2-x1)

(x,y) can either be (1,2) or (4,1); both work

Substitute and solve for b

Find slope, then use slope and one point.

M = ( 1-2)/(4-1) = -1/3

Y - 2 = (-1/3)(X - 1)