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Question Number 54364 by peter frank last updated on 02/Feb/19

$$\mathrm{If}\mathrm{p}\mathrm{is}\mathrm{the}\mathrm{measure}\mathrm{of}\mathrm{per}$$$$\mathrm{pendicular}\mathrm{segment}\mathrm{from}$$$$\mathrm{the}\mathrm{origin}\mathrm{on}\mathrm{the}\mathrm{line}$$$$\mathrm{whose}\mathrm{intercept}\mathrm{are}$$$$\mathrm{a}\mathrm{and}\mathrm{b}.\mathrm{show}\mathrm{that}$$$$\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}=\frac{1}{{\mathrm{p}}^2}$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

$$eqnstline\frac{x}{a}+\frac{y}{b}=1$$$$\mid \frac{\frac{0}{a}+\frac{0}{b}-1}{\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}}\mid =p$$$$\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}}=p^2$$$$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}$$

Commented by peter frank last updated on 03/Feb/19

$$\mathrm{thank}\mathrm{you}\mathrm{mr}\mathrm{tanmay}$$

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