In xy plane the line l has equation y= -x. point p lies on l and has coordinates(-2,2) If l is rotated counterclockwise 45 degree about the origin. what will be the image of p under this rotation.
(-2√2,0)
Just graph the point. See that it is at the end of the diagonal of a square. Now rotate it.
To find the image of point P (-2, 2) after rotating counterclockwise 45 degrees about the origin, we can use the rotation formula.
The rotation formula for a point (x, y) counterclockwise about the origin by angle θ is:
x' = x * cos(θ) - y * sin(θ)
y' = x * sin(θ) + y * cos(θ)
In this case, θ = 45 degrees.
Using the formula, we can substitute the values of P (-2, 2) into the equations:
x' = (-2) * cos(45) - (2) * sin(45)
y' = (-2) * sin(45) + (2) * cos(45)
To find the values of cos(45) and sin(45), we can use their standard values:
cos(45) = sqrt(2)/2
sin(45) = sqrt(2)/2
Substituting these values, we get:
x' = (-2) * (sqrt(2)/2) - (2) * (sqrt(2)/2)
y' = (-2) * (sqrt(2)/2) + (2) * (sqrt(2)/2)
Simplifying further:
x' = -2 * sqrt(2)/2 - 2 * sqrt(2)/2
y' = -2 * sqrt(2)/2 + 2 * sqrt(2)/2
x' = -sqrt(2) - sqrt(2)
y' = -sqrt(2) + sqrt(2)
Combining like terms:
x' = -2sqrt(2)
y' = 0
Therefore, the image of point P (-2, 2) after rotating counterclockwise 45 degrees about the origin is (-2sqrt(2), 0).