Mathematics - Straight Lines Question with Solution | TestHub

$x+y=1$ meets $x$-axis at $A(1,0)$ and $y$-axis at $B(0,1)$. The ordinates of P are ( $1/2,1/2$ ) and ${\mathrm{PP}}_1$ is perpendicular to OA . $\Rightarrow {\mathrm{OP}}_1={\mathrm{P}}_1\mathrm{P}=1/2$

Equation of the line OP is $\mathrm{y}=\mathrm{x}$.

We have ${\left({\mathrm{OM}}_{\mathrm{n}-1}\right)}^2={\left({\mathrm{OP}}_{\mathrm{n}}\right)}^2+{\left({\mathrm{P}}_{\mathrm{n}}{\mathrm{M}}_{\mathrm{n}-1}\right)}^2$ $=2{\left({\mathrm{OP}}_{\mathrm{n}}\right)}^2=2p_n^2$ (say) y B

Also, ${\left({\mathrm{OP}}_{\mathrm{n}-1}\right)}^2={\left({\mathrm{OM}}_{\mathrm{n}-1}\right)}^2+{\left({\mathrm{P}}_{\mathrm{n}-1}{\mathrm{M}}_{\mathrm{n}-1}\right)}^2=2{\mathrm{p}}_{\mathrm{n}}^2+2{\mathrm{p}}_{\mathrm{n}}^2$ $\Rightarrow {\mathrm{p}}_{\mathrm{n}}^2=\frac{1}{4}{\mathrm{p}}_{\mathrm{n}-1}^2\Rightarrow {\mathrm{p}}_{\mathrm{n}}=\frac{1}{2}{\mathrm{p}}_{\mathrm{n}-1}$ $\therefore {\mathrm{OP}}_{\mathrm{n}}={\mathrm{p}}_{\mathrm{n}}=\frac{1}{2}{\mathrm{p}}_{\mathrm{n}-1}=\frac{1}{2^2}{\mathrm{p}}_{\mathrm{n}-2}=\dots =\frac{1}{2^{\mathrm{n}-1}}{\mathrm{p}}_1=\frac{1}{2^{\mathrm{n}}}$