What is the equation of a line that passes through the points (4,8) and (6,1)

Question

What is the equation of a line that passes through the points (4,8) and (6,1)

Answer 1

Hi Foul,

First consider the equation for the slope. m = (y2 - y1)/(x2 - x1) = (1 - 8)/(6 - 4) = -7/2.

Next consider the linear point-slope formula y = mx + b. Choose a point and plug it in (either will do, the result will be the same).

1 = (-7/2)(6) + b = -21 + b -> b = 22 - > y = (-7/2)x + 22 is the equation you desire.

I hope this helps!

Rob

Answer 2

Start out with the general equation of a line: y=mx+b.

Plug in the values at the first point (x=4, y=8) to get [8] = m[4]+b

Plug in the values at the second point (x=6, y=1) to get [1] = m[6]+b

Subtract the values in the second equation from the first equation to get

(8-1) = (4m-6m) + (b-b)

Simply to 7=-2m

Divide by the coefficient of m to get -7/2=m

Plug the value of m into the first equation to get 8=(-7/2)(4)+b

Simply to 8=-14+b

Add 14 to both sides to get 22=b

You now know both m and b, so the equation of the line is y= (-7/2)x + 22