10. (1, 2) and (3, 12)
x₁ = 1 y₁= 2
x₂ = 3 y₂=12
Using the formula;
m = y₂ - y₁ / x₂ - x₁
= 12 - 2 / 3- 1
= 10 / 2
= 5
y = mx + b
substitute x = 1 y=2 m = 5 in the above and then solve for b
2 = 5(1) + b
2 = 5 + b
2- 5 = b
-3 = b
b = -3
substitute m = 5 and b = -3 in y=mx + b
y = 5x - 3
(6, 2 ) and ( -2, -2)
x₁= 6 y₁= 2 x₂= -2 y₂=-2
m = y₂ - y₁ / x₂ - x₁
m = -2 -2 / -2 -6
= -4 / -8
= 4/8
= 1/2
x = 6 y = 2 m = 1/2
substitute in y = mx + b and then solve for b
2 = 5(1/2) + b
2 - 5/2 = b
4 - 5 / 2 = b
-1 /2 = b
substitute m= - 1/2 and b = -1/2 into y=mx + b
y = -(1/2)x - 1/2
12. (4,1) and (1.4)
x₁= 4 y₁= 1 x₂= 1 y₂= 4
m = y₂ - y₁ / x₂ - x₁
= 4-1 / 1-4
= 3/ -3
= - 1
substitute m = -1 x =4 y= 1
y = mx + b
1 = -1(4) + b
1 = -4 + b
1+4 = b
5 = b
b = 5
substitute m = -1 b=5 into y=mx+b
y= -x + 5