The point P (a, b) undergoes the following three transformations successively :
(a) reflection about the line y=x.
(b) translation through 2 units along the positive direction of x-axis.
(c) rotation through angle $\frac{\pi}{4}$ about the origin in the anti-clockwise direction.
If the co-ordinates of the final position of the point P are $(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}})$, then the value of 2a+b is equal to :
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It's often easier to work backwards from the final point to the initial point by applying the inverse transformations in reverse order. The inverse of rotation by $\theta$ is rotation by $-\theta$. The inverse of translation by $+2$ on the x-axis is translation by $-2$. The inverse of reflection about $y=x$ is the same reflection.
Let's trace the transformations of the point P(a, b).
Step (a): Reflection about the line y=x.
The coordinates are interchanged. The new point P' is (b, a).
Step (b): Translation through 2 units along the positive x-axis.
The x-coordinate increases by 2. The new point P'' is (b+2, a).
Step (c): Rotation through $\theta = \pi/4$ about the origin in the anti-clockwise direction.
Let the coordinates of P'' be $(x, y) = (b+2, a)$.
The new coordinates $(x', y')$ after rotation are given by the formulas:
$x' = x \cos\theta - y \sin\theta$
$y' = x \sin\theta + y \cos\theta$
Here, $\theta = \pi/4$, so $\cos(\pi/4) = \sin(\pi/4) = \frac{1}{\sqrt{2}}$.
The final point P''' is $(x', y')$.
$x' = (b+2)\frac{1}{\sqrt{2}} - a\frac{1}{\sqrt{2}} = \frac{b-a+2}{\sqrt{2}}$.
$y' = (b+2)\frac{1}{\sqrt{2}} + a\frac{1}{\sqrt{2}} = \frac{b+a+2}{\sqrt{2}}$.
We are given that the final coordinates are $(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}})$.
Equating the coordinates:
$\frac{b-a+2}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \implies b-a+2 = -1 \implies b-a = -3$. (Equation 1)
$\frac{b+a+2}{\sqrt{2}} = \frac{7}{\sqrt{2}} \implies b+a+2 = 7 \implies b+a = 5$. (Equation 2)
Now we solve the system of two linear equations for a and b.
Adding Equation 1 and Equation 2:
$(b-a) + (b+a) = -3 + 5$
$2b = 2 \implies b=1$.
Substitute b=1 into Equation 2:
$1+a = 5 \implies a=4$.
So the original point P was (4, 1).
The question asks for the value of 2a + b.
$2a+b = 2(4) + 1 = 8 + 1 = 9$.
