A square of side a lies above the x-axis and has one vertex at the origin. This side passing through the origin makes an angle `alpha(0 lt alpha lt pi//4)` with the positive direction of the x-axis. The equation of its diagonal not passing through the origin is

Options

A. `y(cos alpha + sin alpha) + x(cos alpha-sin alpha)=a`
B. `y(cos alpha - sin alpha) - x(sinalpha-cosalpha)=a`
C. `y(cos alpha + sin alpha) + x(sinalpha-cos alpha)=a`
D. `y(cos alpha + sin alpha) + x(sinalpha-cosalpha)=a`

Answer

A

Explanation

We have,
Slope of OB `=tan ((pi)/(4)+alpha)`
`therefore ` Slope of AC `=-cot((pi)/(4)+alpha)`

`implies ` Slope of AC`=-((cos alpha -sin alpha)/(cos alpha+sin alpha))=(sin alpha-cos alpha)/(sin alpha+cos alpha)`
Thus, the equation of AC is
`y-asin alpha =(sin alpha-cos alpha)/(sin alpha+ cos alpha)(x-a cos alpha)`
`implies -x (sin alpha-cos alpha)+y(sin alpha+ cos alpha)=a`
`implies y (sin alpha+ cos alpha)+x(cos alpha -sin alpha )=a`