math
posted by vicki .
how do you find the equation line of these ordered pairs (1/6, 1/3) and (5/6, 5)?

A common form of a linear equation in the two variables x and y is:
y=mx+b
where m and b designate constants. The origin of the name "linear" comes from the fact that the set of solutions of such an equation forms a straight line in the plane. In this particular equation, the constant m determines the slope or gradient of that line, and the constant term "b" determines the point at which the line crosses the yaxis, otherwise known as the yintercept.
m=(y2y1)/(x2x1)
b=(y1*x2y2*x1)/(x2x1)
In this case:
x1=1/6
x2=5/6
y1= 1/3
y2=5
m=(y2y1)/(x2x1)
m=[5(1/3)]/[(5/6)(1/6)]
m=[5+(1/3)]/(4/6)
m=[(15/3)+(1/3)]/(4/6) Becouse 5=15/3
m(16/3)/(4/6)
m=(16*6)/(3*4)
m=96/12
m=8
b=(y1*x2y2*x1)/(x2x1)
b=[(1/3)*(5/6)((5*(1/6)]/[(5/6)(1/6)]
b=[(5/18)(5/6)]/(4/6)
b=[(5/18(15/18)]/(4/6)
Becouse 5/6=15/18
b=(20/18)/(4/6)
b=(20*6)/(4*18)
b= 120/72
b=(12*10)(12*6)
b=(10/6)
b= 5/3 Becouse (10/6)= 5/3
y=mx+b
y=8x(5/3)