यदि समीकरण y = mx + c तथा `x cos alpha + y sin alpha = p` एक ही सरल रेखा को व्यक्त करें, तो सिद्ध कीजिए कि -
`c = p sqrt(1+m^(2))`.

If line x cos alpha + y sin alpha = p " touches circle " x^(2) + y^(2) =2ax then p =

If the equation 3x + 3y + 5=0 is written in the form x cos alpha + y sin alpha = p , then the value of sin alpha + cos alpha is a) sqrt2 b) (1)/( sqrt2) c) - sqrt2 d) - (1)/( sqrt2)

The lines x cos alpha + y sin alpha = p_1 and x cos beta + y sin beta = p_2 will be perpendicular, if :

The lines x cos alpha + y sin alpha = P_1 and x cos beta + y sin beta = P_2 will be perpendicular, if :

The lines x cos alpha + y sin alpha = P_1 and x cos beta + y sin beta = P_2 will be perpendicular, if :

The lines x cos alpha + y sin alpha = P_1 and x cos beta + y sin beta = P_2 will be perpendicular, if :

The lines x cos alpha + y sin alpha = P_1 and x cos beta + y sin beta = P_2 will be perpendicular, if :

If the equations y=mx+c and x cos alpha+y sin alpha=p represent the same straight line,then p=c sqrt(1+m^(2))(b)c=p sqrt(1+m^(2))cp=sqrt(1+m^(2))(d)p^(2)+c^(2)+m^(2)=1

Find P and alpha by converting equation x+y =1 in xcos alpha + y sin alpha = P form.