Line segment EF has endpoints at E(-2, 1) and F(-5, 5). Line segment GH has endpoints at G(-1, -2) and H(3, 1). What are their lengths?
@jim_thompson5910, can you please help me?
Let's find the length of EF This is the same as finding the distance between points E and F To find the distance between any two points, use the distance formula d = sqrt((x2-x1)^2+(y2-y1)^2) d = sqrt((-5-(-2))^2+( 5- 1)^2) ... Note: (x1,y1) is the first point (-2, 1) and (x2,y2) is the second point (-5, 5). d = sqrt((-5+2)^2+( 5- 1)^2) d = sqrt((-3)^2+(4)^2) d = sqrt(9+16) d = sqrt(25) d = 5 So exact distance between the two points is 5 units. Therefore, EF is 5 units long. I'll let you find the length of GH
I got 5 for GH?
Same here, good job
Thanks :) !
you're welcome
Join our real-time social learning platform and learn together with your friends!