Line segment AB¯¯¯¯¯¯¯¯ has endpoints A(1,8) and B(7,−4). What are the coordinates of the point located 5/6 of the way from A to B?

Answer 1

A(1,8) B(7,-4)

Then point P partitions the line segment in the ratio 5 to 1 = 5:1

Is the formula that divides the line segment with endpoints (x1,y1) and (x2,y2)

into the ratio m to n

(n * x1 + m * x2)/(m+n) (n * y1 + m * y2)/(m+n)

x1 = 1 y1 = 8 x2 = 7 y2 = -4 m=5 n=1

(1*1 + 5*7)/(6) = (1+35)/6 = 36/6 = 6 is the x-coordinate of P

(1*8 + 5*-4)/6 = (8 + -20)/6 = -12/6 = -2 is the y-coordinate of P

P is at (6, -2)

CHECK via Distance formula:

Distance from A to B: sqrt ( (7-1)^2 + (8 - -4)^2)

= sqrt( 6^2 + 12^2)

= sqrt( 36 + 144)

= sqrt( 180) = sqrt( 9*20) = sqrt(9*4*5)

= sqrt( 36*5)

= sqrt(36)*sqrt(5)

= 6*sqrt(5)

Distance from A to P: sqrt ( (6-1)^2 + (8 - -2)^2)

= sqrt( 5^2 + 10^2)

= sqrt( 25 + 100)

= sqrt(125)

= sqrt(25*5)

= sqrt(25)*sqrt(5)

= 5*sqrt(5)

The ratio of AP/AB is 5 * sqrt(5) / [ 6 * sqrt(5)]

= 5/6

P at (6, -2) is the point that divides the line segment

in the ratio 5:1, or 5/6 the distance