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Question

A square of side a lies above the x-axis and has one vertex at the origin. The side passing through it makes an angle α(0<α<π4) with the positive direction of x-axis. The equation of its diagonal not through the origin

A
y(cosαsinα)x(sinαcosα)=a
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B
y(cosα+sinα)+x(sinαcosα)=a
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C
y(cosα+sinα)+x(sinα+cosα)=a
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D
y(cosα+sinα)+x(cosαsinα)=a
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Solution

The correct option is D y(cosα+sinα)+x(cosαsinα)=a
Consider the diagram shown below.
Let line OA makes an angle α with xaxis and OA=a, then coordinates of A are (acosα,asinα).

Also, OBOA. Therefore, OB makes an angle (90+α) with xaxis, then coordinates of B are,
[acos(90+α),asin(90+α)]

Here, the diagonal which is not passing through the origin is AB. Therefore, the equation of this diagonal is,
(yasinα)=acosαasinαasinαacosα(xacosα)
(sinα+cosα)(yasinα)=(sinαcosα)(xacosα)
(sinα+cosα)yasinα(sinα+cosα)=(sinαcosα)
y(sinα+cosα)+x(cosαsinα)=asinα(sinα+cosα)acosα(sinαcosα)
y(sinα+cosα)+x(cosαsinα)=a(sin2α+sinαcosαcosαsinα+cos2α)
y(sinα+cosα)+x(cosαsinα)=a
This is the required equation of the diagonal.
1018676_864628_ans_b14f98035ea14f769a6c89b9519af585.PNG

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