Question

What is the equation of the line in point-slope form containing the points (4, -1) and (10, 0)?

Answer 1

Find slope first
(0+1)/(10-4)= 1/6
Point slope: (y-y1) = m(x-x1)
Solution: (y+1) = 1/6(x-4)

Answer 2

The equation of the line in point-slope will be:

`y\:=\:(1)/(6)x-(5)/(3)`

Explanation:

Given the points

• (4, -1)

• (10, 0)

Finding the slope between (4, -1) and (10, 0)

`\mathrm{Slope}=(y_2-y_1)/(x_2-x_1)`

`\left(x_1,\:y_1\right)=\left(4,\:-1\right),\:\left(x_2,\:y_2\right)=\left(10,\:0\right)`

`m=(0-\left(-1\right))/(10-4)`

`m=(1)/(6)`

We know that the slope-intercept form of the line equation is

`y = mx+b`

where m is the slope and b is the y-intercept

Using the slope-intercept form to find the y-intercept 'b'

`y = mx+b`

substituting m = 1/6 and the point (4, -1)

`-1\:=\:(1)/(6)\left(4\right)+b`

`(2)/(3)+b=-1`

subtract 2/3 from both sides

`(2)/(3)+b-(2)/(3)=-1-(2)/(3)`

`b=-(5)/(3)`

now substituting b = -5/3 and m = 1/6 in the slope-intercept form

`y = mx+b`

`y\:=\:(1)/(6)x+\left(-(5)/(3)\right)`

`y\:=\:(1)/(6)x-(5)/(3)`

Therefore, the equation of the line in point-slope will be:

`y\:=\:(1)/(6)x-(5)/(3)`