JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 1)
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 $${\pi \over 4}$$ about the origin in the anti-clockwise direction.
If the co-ordinates of the final position of the point P are $$\left( { - {1 \over {\sqrt 2 }},{7 \over {\sqrt 2 }}} \right)$$, then the value of 2a + b is equal to :
(a) reflection about the line y = x.
(b) translation through 2 units along the positive direction of x-axis.
(c) rotation through angle $${\pi \over 4}$$ about the origin in the anti-clockwise direction.
If the co-ordinates of the final position of the point P are $$\left( { - {1 \over {\sqrt 2 }},{7 \over {\sqrt 2 }}} \right)$$, then the value of 2a + b is equal to :
13
9
5
7
Explanation
Image of A(a, b) along y = x is B(b, a). Translating it 2 units it becomes C(b + 2, a).
Now, applying rotation theorem
$$ - {1 \over {\sqrt 2 }} + {7 \over {\sqrt 2 }}i = \left( {(b + 2) + ai} \right)\left( {\cos {\pi \over 4} + i\sin {\pi \over 4}} \right)$$
$$ - {1 \over {\sqrt 2 }} + {7 \over {\sqrt 2 }}i = \left( {{{b + 2} \over {\sqrt 2 }} - {a \over {\sqrt 2 }}} \right) + i\left( {{{b + 2} \over {\sqrt 2 }} + {a \over {\sqrt 2 }}} \right)$$
$$\Rightarrow$$ b $$-$$ a + 2 = $$-$$1 ......(i)
and b + 2 + a = 7 ...... (ii)
$$\Rightarrow$$ a = 4; b = 1
$$\Rightarrow$$ 2a + b = 9
Now, applying rotation theorem
$$ - {1 \over {\sqrt 2 }} + {7 \over {\sqrt 2 }}i = \left( {(b + 2) + ai} \right)\left( {\cos {\pi \over 4} + i\sin {\pi \over 4}} \right)$$
$$ - {1 \over {\sqrt 2 }} + {7 \over {\sqrt 2 }}i = \left( {{{b + 2} \over {\sqrt 2 }} - {a \over {\sqrt 2 }}} \right) + i\left( {{{b + 2} \over {\sqrt 2 }} + {a \over {\sqrt 2 }}} \right)$$
$$\Rightarrow$$ b $$-$$ a + 2 = $$-$$1 ......(i)
and b + 2 + a = 7 ...... (ii)
$$\Rightarrow$$ a = 4; b = 1
$$\Rightarrow$$ 2a + b = 9
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