Question

Reflecting the point $(2, - 1)$ about y-axis, coordinate axes are rotated at ${45^0}$ angle in negative direction without shifting the origin after reflection the point is denoted by p and after rotation as q. if the x and y coordinates of p and q are denoted by $X_p$ $Y_p$ and $X_q$ $Y_q$ then find the new coordinates of p.

Answer 1

in reflection of any point when the coordinate axis are at origin reflection means mirror image over the x axis or the y axis. And if the reflection size does not change, shape may or may not change in orientation during the reflection over the y axis y coordinates remain the same and the sign of the x coordinate becomes opposite. If rotating in a negative direction it will rotate the figure clockwise we can detect whether the rotation is clockwise or counterclockwise if it is rotated in positive direction it is counterclockwise if it is in negative direction it is clockwise rotation of angle in negative direction means rotating clockwise.

Formula used: To find the coordinate point after rotation let (X,Y) be the new coordinates and (x,y) be the old coordinates
$X = x\cos \theta + y\sin \theta$
$Y = - x\sin \theta + y\cos \theta$
Here we will place the value of (x,y) coordinates of p that we get after reflection and we put the value of $\theta$

Complete step-by-step answer:
The coordinates of given points are $(2, - 1)$ on reflecting the value of $(2, - 1)$ along the y axis the y coordinate remain same and sign of x coordinate changes
$(x,y) \to ( - x,y)$
$( 2,-1)$ on reflection over y axis will be $( - 2, - 1)$
After reflection we do the rotation of ${45^0}$ in negative direction then we will use the formula of rotation i.e.
$X = x\cos \theta + y\sin \theta$
$Y = - x\sin \theta + y\cos \theta$
Putting the value of (x,y) as $( - 2, - 1)$ in the above formula we will get
$X = x\cos \theta + y\sin \theta$
Substituting the value we get
$X = ( - 2)\cos {45^0} + ( - 1)\sin {45^0}$
Putting the values of $\sin \theta ,\cos \theta$
$\Rightarrow X = \dfrac{{ - 2}}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 2 }} = \dfrac{{ - 3}}{{\sqrt 2 }}$
$Y = - x\sin \theta + y\cos \theta$
Putting the value in this we get
$Y = 2\sin {45^0} - 1\cos {45^0}$
$Y = \dfrac{2}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 2 }}$
$Y = \dfrac{1}{{\sqrt 2 }}$
Final answer: Hence the coordinates of p$\left( {\dfrac{{ - 3}}{{\sqrt 2 }},\dfrac{1}{{\sqrt 2 }}} \right)$

Note: In such type of question we have to break the question we have to break the question in two parts and solve the questions firstly by doing reflection and the rotation in case of rotation it is necessary for the student to understand the way it rotates in clockwise and counterclockwise which defines the sign of coordinates and then by applying formula accordingly