Vectors in the coordinate plane

Question

Find the component AB with the given initial and terminal points.

1. A(2.5,-3), B(-4,1.5)
2. A(1/2,-9), B(6,5/2)

Answer 1

1 distance from A to B is sqr(6.5^2 + 4.5^2) = sqr42.25+20.25) = sqr62.5. = about 7.9

2 A to Be is sqr(5.5^2 + 11.5) = sqr(30.25 + 132.25) = sqr162.5 = about 12.7

the distance is the square root of the sum of squares of the difference in coordinates

for part 1

difference in x coordinates = 4.5

difference in y coordinates = 6.5

4.5 squared = 20.25

6.5 squared =42.25

sum of squares = 62.5

square root of 62.5 = about 7.9

for part 2

x coordinate minus the 2nd x coordinate.

difference in x coordinates = 5.5

then square that difference = 30.25

same with the y coordinates.

y coordinate minus 2nd y coordinate = 11.5

square that = 132.25

then sum those squares = 162.5

then take the square root

sqr162.5 is a little less than 13, as 13^2 = 169

sqr164.5 = about 12.7