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  • The point A (2, 1) is shifted by  unit distance parallel to the line <img a

The point A (2, 1) is shifted by \mathrm{3\sqrt{2}} unit distance parallel to the line \mathrm{x + y = 1} in the direction of increasing ordinate to reach a point B. Find the image of B by the line \mathrm{x + y = 1}.

Option: 1

(-3,1)


Option: 2

(-4,3)


Option: 3

(-2,1)


Option: 4

(-3,2)


Answers (1)

best_answer

Let coordinate of B be (x, y)

\begin{aligned} & \mathrm{ \therefore x=2+r \cos \theta }\\ \\& \mathrm{y=1+r \sin \theta} \end{aligned}

\begin{aligned} & \Rightarrow \text{x}=2+3 \sqrt{2}\left(-\frac{1}{\sqrt{2}}\right) \\ \\& \text{y}=1+3 \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) \end{aligned}

Point B is (-1,4)

\begin{aligned} & \mathrm{B C=\sqrt{2}, B B^{\prime}=2 \sqrt{2}} \\ & \mathrm{\text { Let } B^{\prime} \equiv(h, k), r_1=-2 \sqrt{2}} \end{aligned}

\begin{aligned} & \therefore \mathrm{h}=-1+2 \sqrt{2} \cos \alpha \\ & \mathrm{k}=4+2 \sqrt{2} \sin \alpha \end{aligned}

\Rightarrow \mathrm{h}=-3, \mathrm{k}=2

\mathrm{\therefore \; coordinates\; of \; B^{\prime}\; is (-3,2).}

Here \tan \theta =-1

\begin{aligned} & \therefore \sin \theta=\frac{1}{\sqrt{2}} \\ & \cos \theta=-\frac{1}{\sqrt{2}} \end{aligned}

for increasing ordinate

Here \tan \alpha=1

\begin{aligned} & \Rightarrow \sin \alpha=\frac{1}{\sqrt{2}} \\ & \cos \alpha=\frac{1}{\sqrt{2}} \end{aligned}

 

Posted by

Kuldeep Maurya

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