Question 369580
Find the point that is one-fourth of the way from(2,4) to (10,8).


distance d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )
d = sqrt( (10 - 2)^2 + (8 - 4)^2 )
d = sqrt( 8^2 + 4^2 )
d = sqrt( 64 + 16 )
d = sqrt(80)
80 = 16 * 5
d = 4sqrt(5)
one-fourth of this distance is sqrt(5)
x: 2 + 1/4(10 - 2) = 2 + 1/4(8) = 2 + 2 = 4
(the 10 - 2 is subtracting the x coordinates)
y: 4 + 1/4(8 - 4) = 4 + 1/4(4) = 4 + 1 = 5
(the 8 - 4 is subtracting the y coordinates)
the point (4,5) should be sqrt(5) away from (2,4)
check:
d = sqrt( (4 - 2)^2 + (5 - 4)^2 )
d = sqrt( 2^2 + 1^2)
d = sqrt( 4 + 1 )
d = sqrt(5)
yes, answer is the point (4,5)