The coordinates of Point A are (6,3). Point B is a reflection of Point A across the y axis. In which Quadrant is point B located
?
1 answer:
point B is located in quadrant II (option B)
Explanation:Point A: (6, 3) = (x, y)
reflection of point A across the y axis:
The x coordinate will be negative and the y coordinate will remain the same
reflection of point A across the y axis = (-6, 3)
Attaching the graph showing the point after reflection:
Since point B is the reflection of point A.
Hence, point B is located in quadrant II (option B)
You might be interested in
Answer:
25/2
Step-by-step explanation:
Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]
Where P, Q are scalar functions
We want to compute
Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.
a) Directly
Let us break down C into 4 paths which represents the sides of the rectangle.
is the line segment from (0,0) to (5,0)
is the line segment from (5,0) to (5,1)
is the line segment from (5,1) to (0,1)
is the line segment from (0,1) to (0,0)
Then
Given 2 points P, Q we can always parametrize the line segment from P to Q with
r(t) = tQ + (1-t)P for 0≤ t≤ 1
Let us compute the first integral. We parametrize as
r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and
r'(t) = (5,0) so
Now the second integral. We parametrize as
r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and
r'(t) = (0,1) so
The third integral. We parametrize as
r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and
r'(t) = (-5,0) so
The fourth integral. We parametrize as
r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and
r'(t) = (0,-1) so
So
Now, let us compute the value using Green's theorem.
According with this theorem
where A is the interior of the rectangle.
so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}
We have
so