Question Number 54364 by peter frank last updated on 02/Feb/19
$$\mathrm{If}\:\:\mathrm{p}\:\mathrm{is}\:\mathrm{the}\:\mathrm{measure}\:\mathrm{of}\:\mathrm{per} \\ $$$$\mathrm{pendicular}\:\mathrm{segment}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{origin}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{whose}\:\mathrm{intercept}\:\mathrm{are} \\ $$$$\mathrm{a}\:\:\mathrm{and}\:\:\:\:\mathrm{b}.\mathrm{show}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{p}^{\mathrm{2}} } \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
$${eqn}\:{st}\:{line}\:\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$$\mid\frac{\frac{\mathrm{0}}{{a}}+\frac{\mathrm{0}}{{b}}−\mathrm{1}}{\sqrt{\left(\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} }}\mid={p} \\ $$$$\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}={p}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{{p}^{\mathrm{2}} } \\ $$
Commented by peter frank last updated on 03/Feb/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mr}\:\mathrm{tanmay} \\ $$