The line segment joining A (3, 0), B(5, 2) is rotated about a point A in anticlockwise sense through an angle π4and B move to C. If a point D be the reflection of C in y-axis then D=

Given A (3, 0) B(5, 2)Slope of AB=2−05−3=1∴ AB¯ making angle π4 with x−axisLet θ= Angle made by AC¯ with x−axisθ=450+450=900r=AC¯=AB¯=5−32+2−02=4+4=22∴ c=x1+rcosθ, y1+rsinθ=3+22cos900,0+22sin900=3+0, 0+22=3, 22∈ Q1Reflection of C w.r.to y−axis D∈Q2∴ D=−3, 22

The line segment joining A (3, 0), B(5, 2) is rotated about a point A in anticlockwise sense through an angle π4and B move to C. If a point D be the reflection of C in y-axis then D=

Given A (3, 0) B(5, 2)Slope of AB=2−05−3=1∴ AB¯ making angle π4 with x−axisLet θ= Angle made by AC¯ with x−axisθ=450+450=900r=AC¯=AB¯=5−32+2−02=4+4=22∴ c=x1+rcosθ, y1+rsinθ=3+22cos900,0+22sin900=3+0, 0+22=3, 22∈ Q1Reflection of C w.r.to y−axis D∈Q2∴ D=−3, 22

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The line segment joining A (3, 0), B(5, 2) is rotated about a point A in anticlockwise sense through an angle $\frac{π}{4} a n d B$ move to C. If a point D be the reflection of C in y-axis then D=

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$(3, 8 \sqrt{2})$

$(3, 2 \sqrt{2})$

$(3, - 2 \sqrt{2})$

$(- 3, 2 \sqrt{2})$

answer is A.

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Detailed Solution

Given A (3, 0) B(5, 2)

$\begin{array}{l} S l o p e o f A B = \frac{2 - 0}{5 - 3} = 1 \\ ∴ \bar{A B} m a k i n g a n g l e \frac{π}{4} w i t h x - a x i s \\ L e t θ = A n g l e m a d e b y \bar{A C} w i t h x - a x i s \\ θ = 45^{0} + 45^{0} = 90^{0} \\ r = \bar{A C} = \bar{A B} = \sqrt{{(5 - 3)}^{2} + {(2 - 0)}^{2}} = \sqrt{4 + 4} = 2 \sqrt{2} \\ ∴ c = (x_{1} + r \cos θ, y_{1} + r \sin θ) = (3 + 2 \sqrt{2} \cos 90^{0}, 0 + 2 \sqrt{2} \sin 90^{0}) \\ = (3 + 0, 0 + 2 \sqrt{2}) \\ = (3, 2 \sqrt{2}) \in Q_{1} \\ Re f l e c t i o n o f C w . r . t o y - a x i s D \in Q_{2} \\ ∴ D = (- 3, 2 \sqrt{2}) \end{array}$