Question

One side of a square of length $a$ is inclined to the positive x-axis at an angle $\alpha$ with one of the vertices of the square at the origin. The equation of a diagonal of the square is

• A. $y(\cos \alpha -\sin \alpha )=x(\cos \alpha +\sin \alpha )$
• B. $y(\cos \alpha +\sin \alpha )=x(\cos \alpha -\sin \alpha )$
• C. $y(\sin \alpha +\cos \alpha )-x(\sin \alpha -\cos \alpha )=a$
• D. $y(\sin \alpha +\cos \alpha )+x(\sin \alpha -\cos \alpha )=a$

Answer 1

Answer: (A), (C)

$y(\cos \alpha -\sin \alpha )=x(\cos \alpha +\sin \alpha )$
$y(\sin \alpha +\cos \alpha )-x(\sin \alpha -\cos \alpha )=a$

If we assume that AB is inclined to x-axis like in figure 1, with an angle $\alpha$ then $\angle CAM=45+\alpha$
Coordinates of C are given by C($\sqrt{2}a\cos (45+\alpha ),\sqrt{2}a\sin (45+\alpha )$)
Equation of AC is given by,
$y=mx,wherem=\tan (45+\alpha )y=\frac{1+\tan \alpha }{1-\tan \alpha }xy(\cos \alpha -\sin \alpha )=(\sin \alpha +\cos \alpha )x$
If we assume AB is inclined with x-axis like in figure 2, with an angle $\alpha$ then $\angle CAM=\alpha -45$
Coordinates of C are given by C($\sqrt{2}a\cos (\alpha -45),\sqrt{2}a\sin (\alpha -45)$)
Then, equation of AC will be $y(\sin \alpha +\cos \alpha )=(\sin \alpha -\cos \alpha )x$
Similarly, solve for diagonals BD for both figures.
In figure 1, we get
B($a\cos \alpha ,a\sin \alpha$) D($-a\sin \alpha ,a\cos \alpha$)
Using two points form, we will get equation of BD as
$y-a\sin \alpha =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }(x-a\cos \alpha )y(\sin \alpha +\cos \alpha )-x(\sin \alpha -\cos \alpha )=a$
similarly for figure 2, we get equation of BD as
$y-a\sin \alpha =\frac{\sin \alpha +\cos \alpha }{\cos \alpha -\sin \alpha }(x-a\cos \alpha )y(\cos \alpha -\sin \alpha )-x(\sin \alpha +\cos \alpha )+a=0$