1.

If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that `1/(p^2)=1/(a^2)+1/(b^2)`.

Answer» The equation of the line making intercepts a and b on the axes is given by `(x)/(a)+(y)/(b)=1, i.e. (x)/(a)+(y)/(b)=1=0.....(i)`
Since p is the length of the perpendicular from O(0,0) to line (i), we have `p=(|(1)/(a)xx0+(1)/(b)xx0-1|)/(sqrt((1)/(a^(2))+(1)/(b^(2))))=(|-1|)/(sqrt((1)/(a^(2))+(1)/(b^(2))))=-(1)/(sqrt((1)/(a^(2))+(1)/(b^(2))))`
`Rightarrow p^(2)=(1)/(((1)/(a^(2))+(1)/(b^(2))))=(a^(2)b^(2))/((b^(2)+a^(2)))`
`Rightarrow (1)/(p^(2))=((b^(2)+a^(2)))/(a^(2)b^(2))=((b^(2))/(a^(2)b^(2))+(a^(2))/(a^(2)b^(2)))=((1)/(a^(2))+(1)/(b^(2)))`
`"Hence"(1)/(p^(2))=(1)/(a^(2))+(1)/(b^(2))`


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