Show that the equation of the straight line `xcosalpha + y sin alpha = p` can be expressed in the following form `(x-pcos alpha)/(-sin alpha)=(y-psin alpha)/(cos alpha)=r`

Show that the equation of the straight line xcosalpha+ysinalpha=p can be expressed in the following form: (x-pcosalpha)/(-sinalpha)=(y-p sin alpha)/(cosalpha)=r

Reduce the equation of the line x cos alpha+y sin alpha-p=0 into intercepts form

Show that the image of the point (h,k) with respect to the striaight line x cos alpha+ y sin alpha=p is the point (2 p cos alpha- h cos2 alpha- k sin 2 alpha, 2p sin alpha- h sin 2 alpha- k cos 2 alpha) .

The angle made by the straight line x cos alpha + y sin alpha = p with the negative direction of x - axis is _

Show that the straight line x cos alpha+y sin alpha=p touches the curve xy=a^(2), if p^(2)=4a^(2)cos alpha sin alpha

if the equations y=mx+c and x cos alpha+y sin alpha=p represent the same straight line then:

The line x cos alpha+y sin alpha=p touches the circle x^(2)+y^(2)-2ax cos alpha-2ay sin alpha=0. then p=

The area of the triangle formed by the line x cos alpha+y sin alpha=p with the coordinate axes is