What is the image point of left bracket, minus, 1, comma, 1, right bracket(−1,1) after a translation right 2 units and down 1 unit?
The image point of (-1,1) after a translation right 2 units and down 1 unit would be (1,0).
The image point of (5,2) after a translation left 1 unit and down 2 units would be (4,0).
What is the image point of left bracket, 6, comma, 8, right bracket(6,8) after a translation right 5 units and up 3 units?
The image point of (6,8) after a translation right 5 units and up 3 units would be (11, 11).
What is the image point of left bracket, 0, comma, 4, right bracket(0,4) after a translation right 2 units and down 3 units?
The image point of (0,4) after a translation right 2 units and down 3 units would be (2,1).
What is the image of left bracket, 1, comma, 6, right bracket(1,6) after a reflection over the y-axis?
The image of (1,6) after a reflection over the y-axis would be (-1, 6).
What is the image of left bracket, minus, 1, comma, 5, right bracket(−1,5) after a reflection over the line y, equals, xy=x?
To find the image of a point after a reflection over a line given by the equation y = f(x), we can find the equation of the perpendicular line to f(x) passing through the point. Then we find the intersection point of the two lines, and the image point will be the reflection of the original point over this intersection point.
The equation xy = x can be rearranged as y = 1/x.
The slope of y = 1/x is -1 (by taking the derivative of y = 1/x), so the slope of the perpendicular line is 1.
The equation of the perpendicular line passing through (-1, 5) is y - 5 = 1(x - (-1)), which simplifies to y = x + 6.
To find the intersection point of y = 1/x and y = x + 6, we can substitute y = x + 6 into the equation y = 1/x:
x + 6 = 1/x
(x + 6)x = 1
x^2 + 6x - 1 = 0
Solving this quadratic equation will give us the x-coordinate of the intersection point. Using the quadratic formula, we have:
x = (-6 ± √(6^2 - 4(1)(-1))) / 2
x = (-6 ± √(36 + 4)) / 2
x = (-6 ± √40) / 2
x = (-6 ± 2√10) / 2
x = -3 ± √10
Taking the positive root, x = -3 + √10.
Substituting this back into y = 1/x, we have:
y = 1/(-3 + √10)
y = 1/(-3 + √10) * (-3 - √10)/(-3 - √10)
y = (-3 - √10)/(-9 + 10)
y = (-3 - √10)/1
y = -3 - √10
So the intersection point is (-3 + √10, -3 - √10).
The image point of (-1, 5) after a reflection over the line y = xy = x is the reflection of (-1, 5) over (-3 + √10, -3 - √10).
To find this reflection point, we can calculate the coordinates of the midpoint between (-1,5) and (-3 + √10, -3 - √10):
x-coordinate of midpoint = (x1 + x2) / 2 = (-1 + (-3 + √10)) / 2 = -2 + √10/2
y-coordinate of midpoint = (y1 + y2) / 2 = (5 + (-3 - √10)) / 2 = 1 - √10/2
So the reflection point is (-2 + √10/2, 1 - √10/2).
Finally, the image point of (-1, 5) after the reflection is the reflection of (-1, 5) over (-2 + √10/2, 1 - √10/2), which is:
(2((-2 + √10)/2) - (-1), 2(1 - √10/2) - 5)
Simplifying, we get:
((√10 - 2), -√10 - 1)