if `P` is the length of perpendicular from origin to the line `x/a+y/b=1` then prove that `1/(a^2)+1/(b^2)=1/(p^2)`

The length of perpendicular from the origin on the line x/a + y/b = 1 is -

(i ) If the length of perpendicular from origin to the line ax+by+a+b=0 is p , then show that : p^(2)-1=(2ab)/(a^(2)+b^(2)) (ii) If the length of perpendicular from point (1,1) to the line ax-by+c=0 is unity then show that : (1)/(a)-(1)/(b)+(1)/(C )=(c )/(2ab)

If p be the length of theperpendicular from the origin on the line x//a+y//b =1, then p^2=a^2+b^2 b. p^2=1/(a^2)+1/(b^2) c. 1/(p^2)=1/(a^2)+1/(b^2) d. none of these

If 2p is the length of perpendicular from the origin to the lines (x)/(a)+(y)/(b)=1, then a^(2),8p^(2),b^(2) are in

If p is the length of perpendicular from point (1,1) to the straight line ax+by+a+b=0 , then prove that : p^(2)=4+(8ab)/(a^(2)+b^(2))

Consider the following statements : 1. The length p of the perpendicular from the origin to the line ax+by=c satisfies the relation p^(2)=c^(2)/(a^(2)+b^(2)) The length p of the pependicular from the origin to the line x/a+y/b=1 satisfies the relation 1/p^(2)=1/a^(2)+1/b^(2) . 3. The length p of the perpendicular from the origin to the line y=mx+c satisfies the relation 1/p^(2)=(1+m^(2)+c^(2))/c^(2) Which of the above is/are correct ?

If p is the length of perpendicular from the origin on the line (x)/(a)+(y)/(b)=1 and a^(2) , p^(2) and b^(2) are in AP, the show that a^(4)+b^(4)=0 .

I.If p is the length of the perpendicular from the origin on the line (x)/(a)+(y)/(b)=1 and a^(2),p^(2),b^(2) are in A.P,then a^(4)-2p^(2)a^(2)+2p^(4)=