Question 1204564

Let's Assume That The Given Line is {{{AB}}}, and the point is {{{P}}}, which is not on {{{AB}}}.

Now, Let's assume, We have drawn a perpendicular {{{BV}}} on {{{AB}}}, and intersecting line {{{AB}}} at a point{{{ O}}}.

We have to prove that, this {{{BV}}} is the {{{only}}} line passing through {{{P}}} that is {{{perpendicular }}}to {{{AB}}}.

Now, we will use a construction.

{{{ drawing( 600, 600, -10, 10, -10, 10, 
blue(arrow(1,2,8,2)),blue(arrow(8,2,1,2)),
locate(1.3,2.1,A),locate(4,8,B),locate(4,2.1,B),
locate(4,6,P),circle(4,6,.12),
red(arrow(4,8.5,4,-8.5)),red(arrow(4,-8.5,4,8.5)),
locate(4,2,O),green(arrow(4,6,7,-7)),
locate(4,5,F),locate(6,-5,G),locate(4,-5,V),
graph( 600, 600, -10, 10, -10, 10, 0)) }}}


Let's construct another perpendicular{{{ FG }}}on {{{AB}}} from point {{{P}}}.

 Proof:

We have,

{{{BV }}}perpendicular {{{AB}}}. 

 Also, {{{FG}}} perpendicular {{{AB}}}.

So, {{{BV}}} || {{{FG}}}. [Both are perpendiculars on the same line.]

Now since both {{{BV}}} and {{{FG}}} have point {{{P }}in common and they are parallel, means they should {{{coincide}}}.

So, {{{BV}}} and {{{FG }}}are the coincident lines or parallel lines and only one of them passes through point a{{{P}}}}


answer:


A.{{{BV}}} and {{{FG}}} cannot both be perpendicular to  {{{AB}}}