# Find the point 1/3 of the way from A(2,5) to B(8, -1) Since we are working with a line, we can look at 1/3 the distance between the x points, and 1/3 the distance between the y points.

We have 2 and 8 for our x values, for a distance of 6 units between them.  1/3 of 6 is 2.  So the...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Since we are working with a line, we can look at 1/3 the distance between the x points, and 1/3 the distance between the y points.

We have 2 and 8 for our x values, for a distance of 6 units between them.  1/3 of 6 is 2.  So the x value of our new point is 2 units away from A.

We have -1 and 5 for our y values, a distance of 6 between them.  1/3 of 6 is 2.  So the y value of our new point is 2 units away from A. *note* that the y values are decreasing.

Our new point is (2 + 2, 5 + 2) = (4, 3)

Approved by eNotes Editorial Team Find the point 1/3 of the way from A(2,5) to B(8,-1):

These points lie on the line y=-x+7.

(The slope is m=(5-(-1))/(2-8)=-1. Use the point-slope form to get:

y-5=-1(x-2) or y=-x+7.)

The distance from A to B is `6sqrt(2) ` . Use the distance formula:

`D=sqrt((8-2)^2+(-1-5)^2)=sqrt(72)=6sqrt(2) `

The point 1/3 from A to B lies on the line and is `2sqrt(2) ` units away from A.

From A, go over two units and down 2 units to point P(4,3). The distance from A to P is `2sqrt(2) ` as needed, and P lies on the line AB as the slope from A to P is -1.

---------------------------------------------------------------------------

The required point is (4,3)

--------------------------------------------------------------------------

Note that you can also see that to get from A to B we go 6 units right (parallel to the horizontal axis) and 6 units down (parallel to the vertical axis), so the point 1/3 of the way is right 2 down 2 from A.

The graph: