Posted by **vicki** on Friday, February 11, 2011 at 4:08pm.

how do you find the equation line of these ordered pairs (1/6, -1/3) and (5/6, 5)?

- math -
**Bosnian**, Friday, February 11, 2011 at 10:50pm
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 y-axis, otherwise known as the y-intercept.

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

b=(y1*x2-y2*x1)/(x2-x1)

In this case:

x1=1/6

x2=5/6

y1= -1/3

y2=5

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

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*x2-y2*x1)/(x2-x1)

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)

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