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  • A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P.

A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P.

Answers (1)

Let the variable line be ax+by=1  

  Length of perpendicular from (2,0) to the line ax+by-1=0 

 d=\frac{\left | 2*a+0*b-1 \right |}{\sqrt{a^{2}+b^{2}}}=\frac{2a-1}{\sqrt{a^{2}+b^{2}}}

Now perpendicular distance from B(0,2) = \left | \frac{0*a+2*b-1}{\sqrt{a^{2}+b^{2}}} \right |

  Now, perpendicular distance from C(1,1)= \left | \frac{1*a+1*b-1}{\sqrt{a^{2}+b^{2}}} \right |

 The algebraic sum of the perpendicular from the given points (2,0), (0,2) 

and (1,1) to this line is zero. 

 d1+d2+d3=0  

 \frac{2a-1}{\sqrt{a^{2}+b^{2}}}+\frac{2b-1}{\sqrt{a^{2}+b^{2}}}+\frac{a+b-1}{\sqrt{a^{2}+b^{2}}}=0

 2a-1+2b-1+a+b-1=0 

  3a+3b-3=0  

  a+b-1=0

   a+b=1

  So, the equation ax+by=1 represents a family of straight lines passing

 through a fixed point .  

  Comparing equation ax+by=1 and a+b=1, we get x=1 and y=1. 

 Coordinates of fixed point is (1,1)

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