Find the coordinates of the point 3/10 of the way from A to B

Question

the coordinates are (-3,6) and (11,5)

Answer 1

First let's find the distance using distance formula:

A(-3,6) and B(11,5)

d = √((-3-11)² + (6-5)²)

d =√(-14)² +1²)

d = √(196 +1)

d = √197

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Therefore:

(3/10)d = (3/10)√197

The slope the line is:

m = (6-5)/(-3-11) = 1/(-14) = -1/14

The equation of the line is:

y-6 = -1/14 (x+3)

(x,y) = to the coordinate of 3/10 away from point A to B

(3/10)√197 = √((x+3)²+(y-6)²)

Using this and the equation of the line, we can solve for the value of (x,y) by systems of nonlinear equations.

The two solutions are ( 6/5, 57/10) and (-36/5,65/10). We eliminate (-36/5,65/10) because it's not going toward B.

Therefore the answer to the problem is (6/5,57/10)

Answer 2

A(-3, 6) and B(11,5) have the point (1.2, 5.7)

= (6/5, 57/10) 3/10 of the way from A to B

11 - - 3 = 11+3 = 14. 14 x 3/10 = 42/10 = 4.2 -3+4.2 = 1.2 = the x coordinate of the point 3/10 of the way from A to B

6-5 = 1. 1 x 3/10 = 0.3. 6 - .3 = 5.7 = the y coordinate of the point 3/10 of the way from A to B

Since you listed (-3,6) first, it's likely point A

But if (11,5) were A and (-3,6) were B, then the point 3/10 of the way from A to B would be (6.8, 5.3) = (34/5, 53/10)

14 x 3/10 = 4.2. 11-4.2 = 6.8

1 x 3/10 = 0.3 5+.3 = 5.3