We know that diagonal of a square makes an angle of 45° with its sides.

So, \(\angle DCA = 45°\)

Therefore by linear pair.

Given that the square has a side ‘a’.

Thus the coordinates of the vertex C will be\( (a cosα, a sinα).\)

By applying angle sum property to triangle DCE, we get;

\(\angle CED = 45^{\circ} – \alpha \)

Hence slope of the diagonal is – \(tan (45^{\circ} – \alpha)\) \(\Rightarrow -\frac{1 – tan \alpha}{1 + tan \alpha}\) \(\Rightarrow -\frac{cos \alpha – sin\alpha}{cos \alpha + sin\alpha}\)

Since it passes through C, we get the equation as

\(\Rightarrow \frac{y – a sin \alpha }{x – a cos \alpha } = – (\frac{cos \alpha – sin \alpha}{cos \alpha + sin })\) \(y(cos \alpha + sin \alpha) – a \;cos\alpha \;sin \alpha – a \;sin^{2}\alpha = – x (cos \alpha – sin \alpha) + a \;cos^{2}\alpha – a\; sin \alpha\; cos \alpha\) \(\Rightarrow y(sin \alpha + cos \alpha) + x (cos \alpha – sin \alpha) = a (cos^{2}\alpha + sin^{2}\alpha)\) \(\Rightarrow y (sin \alpha + cos \alpha) + x (cos \alpha – sin \alpha) = a\)

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