Question #5d6a1

Question

Question #5d6a1

1 Answer

Related questions

A line segment has endpoints at $(5, 8)$ $(5 , 8)$ and $(6, 1)$ $(6 , 1)$ . If the line segment is rotated about...
A line segment has endpoints at $(5, 2)$ $(5 , 2)$ and $(3, 1)$ $(3 , 1)$ . If the line segment is rotated about...
A line segment has endpoints at $(5, 2)$ $(5 , 2)$ and $(3, 1)$ $(3 , 1)$ . If the line segment is rotated about...
A line segment has endpoints at $(8, 4)$ $(8 , 4)$ and $(1, 2)$ $(1 , 2)$ . If the line segment is rotated about...
A line segment has endpoints at $(8, 4)$ $(8 , 4)$ and $(1, 2)$ $(1 , 2)$ . If the line segment is rotated about...
A line segment has endpoints at $(2, 3)$ $(2 , 3)$ and $(1, 2)$ $(1 , 2)$ . If the line segment is rotated about...
A line segment has endpoints at $(2, 3)$ $(2 , 3)$ and $(1, 2)$ $(1 , 2)$ . If the line segment is rotated about...
A line segment has endpoints at $(2, 3)$ $(2 , 3)$ and $(5, 2)$ $(5 , 2)$ . If the line segment is rotated about...
A line segment has endpoints at $(2, 2)$ $(2 , 2)$ and $(5, 4)$ $(5 , 4)$ . If the line segment is rotated about...
A line segment has endpoints at $(2, 2)$ $(2 , 2)$ and $(5, 4)$ $(5 , 4)$ . If the line segment is rotated about...

Impact of this question

1686 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-20T23:06:08

$= P (- 1, 0), L (0, 1), A (1, 1), T (0, 0)$ $= P(-1,0), L(0,1), A(1,1), T(0,0)$

Explanation:

A translation of $(x + 1, y)$ $(x+1,y)$ means for all cooridnates stated, move 1 unit forward on the $x$ $x$ axis, and 0 units forward on the $y$ $y$ axis. The cooridnates given are $P (- 2, 0), L (- 1, 1), A (0, 1), T (- 1, 0)$ $P(-2,0),L(-1,1),A(0,1),T (-1,0)$ . If I add 1 on the $x$ $x$ -axis of $P$ $P$ , I get

$(- 2 + 1, 0) = (- 1, 0)$ $(-2+1,0)=(-1,0)$ , therefore $(- 1, 0)$ $(-1,0)$ is the new $P$ $P$ coordinate. I then do the same thing to
$L (- 1 + 1, 1) = (0, 1)$ $L (-1+1,1)=(0,1)$
$A (0 + 1, 1) = (1, 1)$ $A (0+1,1)=(1,1)$ , and
$T (- 1 + 1, 0) = (0, 0)$ $T (-1+1,0)=(0,0)$ .

Therfore, the new cooridnates of PLAT are $P (- 1, 0), L (0, 1), A (1, 1) and T (0, 0$ $P(-1,0), L(0,1), A(1,1) and T(0,0$ . These new coordinates are the same shape, only shifted 1 unit forward in the $x$ $x$ -direction.

I hope I helped!

Science

Math

Humanities

... and beyond

Question #5d6a1

1 Answer

Explanation:

Related questions

Impact of this question