Sol:

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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