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Two fixed straight lines \mathrm{x-axis \: and\: y=m x} are cut by a variable line in the points \mathrm{A(a, 0) \: and \: B(b, m b)} respectively. \mathrm{P \: and\: Q} are the feet of the perpendiculars drawn from  \mathrm{A\: and \: B} upon the lines \mathrm{y=m x\: and\: x-}axis. Show that, if \mathrm{A B} passes through a fixed point \mathrm{ (h, k)}, then \mathrm{P Q}  will also pass through a fixed point. Find the fixed point.
 

Option: 1

\mathrm{ \left(\frac{h+m k}{1+m^2}, \frac{m h-k}{1+m^2}\right)}
 


Option: 2

\mathrm{\left(\frac{h-m k}{1+m^2}, \frac{m h+k}{1+m^2}\right)}
 


Option: 3

\mathrm{\left(\frac{h+m k}{1-m^2}, \frac{m h-k}{1-m^2}\right)}
 


Option: 4

\mathrm{\left(\frac{h-m k}{m^2-1}, \frac{m h+k}{m^2-1}\right)}


Answers (1)

best_answer

\mathrm{A}, \mathrm{R}$ and $\mathrm{B} are collinear

then \mathrm{\quad \frac{k-0}{h-a}=\frac{m b-0}{b-a}}

\therefore \quad \frac{\mathrm{a}}{\mathrm{b}} \mathrm{k}-\mathrm{am}+\mathrm{mh}-\mathrm{k}=0  ..............(i)

equation of \mathrm{AP} is

\mathrm{ y-0=-\frac{1}{m}(x-a) }
co-ordinate of \mathrm{ \mathrm{P} } is \left(\frac{\mathrm{a}}{1+\mathrm{m}^2}, \frac{\mathrm{am}}{1+\mathrm{m}^2}\right)

\therefore equation of \mathrm{PQ} is

\mathrm{ \frac{a}{b}(y-m x)+a m-\left(1+m^2\right) y=0}
or \mathrm{\quad-\frac{a}{b}(y-m x)-a m+\left(1+m^2\right) y=0}     ............(ii)
Comparing equation (i) and (ii) we get

\mathrm{ y-m x=-k,\left(1+m^2\right) y=m h-k }

\mathrm{\Rightarrow \left(1+m^2\right)(m x-k)=m h-k }

\mathrm{\Rightarrow x=\frac{h+m k}{1+m^2} \text { and } y=\frac{m h-k}{1+m^2}}

Hence line PQ pass through a fixed point

\left(\frac{\mathrm{h}+\mathrm{mk}}{1+\mathrm{m}^2}, \frac{\mathrm{mh}-\mathrm{k}}{1+\mathrm{m}^2}\right)

\quad(\because \mathrm{m}, \mathrm{h} \text { and } \mathrm{k} \text { are fixed })

Hence option 1 is correct.

 

Posted by

Devendra Khairwa

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