Just use the good old Pythagorean Theorem:
d^2 = ∆x^2 + ∆y^2
= ((3x+1)^2-(2x)^2)+((y-2)-(y-2))^2
= (5x^2 + 6x + 1) - (0)
. . .
Determine the length of the line segment between the following points.
T(2x;y-2) and U(3x+1;y-2)
5 answers
Still confused steve can you please elaborate more PLEASE
what Steve meant to say was:
d^2 = (3x+1 - 2x)^2 + (y-2 - (y-2))^2
= (x+1)^2 + 0^2
d =√(x+1)^2
= x+1
e.g.
let x = 3, y = 5
then T is (6,3) and U is (10,3)
TU = √(4^2 + 0^2)
= √16 = 4
according to my result, d = 3+1 = 4 , as in the test answer.
d^2 = (3x+1 - 2x)^2 + (y-2 - (y-2))^2
= (x+1)^2 + 0^2
d =√(x+1)^2
= x+1
e.g.
let x = 3, y = 5
then T is (6,3) and U is (10,3)
TU = √(4^2 + 0^2)
= √16 = 4
according to my result, d = 3+1 = 4 , as in the test answer.
ouch! Guess I had too many squares working with me!
What was I thinking?
What was I thinking?
Thank u guys ...I now understand
1 answer