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  • 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

Q : 18    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  \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}.

Answers (1)

best_answer

we know that intercept form of line is
\frac{x}{a}+\frac{y}{b} = 1
we know that
d = \left | \frac{Ax_1+bx_2+C}{\sqrt{A^2+B^2}} \right |
In this problem
A = \frac{1}{a},B = \frac{1}{b}, C =-1 \ and \ (x_1,y_1)= (0,0)
p= \left | \frac{\frac{1}{a}\times 0+\frac{1}{b}\times 0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \right | = \left | \frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \right |
On squaring both the sides 
we will get
\frac{1}{p^2}= \frac{1}{a^2}+\frac{1}{b^2}
Hence proved

Posted by

Gautam harsolia

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