Determine the length of the line segment between the following points.
T(2x;y-2) and U(3x+1;y-2)
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)
. . .
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.
ouch! Guess I had too many squares working with me!
What was I thinking?
Thank u guys ...I now understand
To determine the length of the line segment between points T(2x, y-2) and U(3x+1, y-2), we can use the distance formula.
The distance formula states that the distance between two points in a coordinate plane, (x1, y1) and (x2, y2), is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, the coordinates of point T are (2x, y-2), and the coordinates of point U are (3x+1, y-2).
We can substitute the coordinates into the distance formula to find the length of the line segment:
d = √((3x+1 - 2x)^2 + ((y-2) - (y-2))^2)
Simplifying this expression:
d = √((x+1)^2 + 0^2)
The second term simplifies to 0 because (y-2) - (y-2) equals 0.
Now, we can simplify the expression further:
d = √((x^2 + 2x + 1) + 0)
The expression (x^2 + 2x + 1) can be factored as (x + 1)(x + 1), which simplifies to (x + 1)^2.
Thus, the length of the line segment between points T(2x, y-2) and U(3x+1, y-2) is:
d = √((x + 1)^2)
Therefore, the length of the line segment is |x + 1|.