Question

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.​

Answer 1

Answer:

P ( 1 , 1 ).

Step-by-step explanation:

Let P be ( x₁ , y₁ ) line is y - y₁ = m ( x - x₁ )

= > m x - y ( y₁ - m x₁ ) = 0 ... ( i )

= > y₁ - m x₁ = y - m x ... ( ii )

Sum of perpendicular distance from ( 2 , 0 ) , ( 0 , 2 ) , ( 1 , 1 ) on ( i ) = 0

= > ( 2 m + y₁ - m x₁ ) / √ ( m² + 1 ) + ( - 2 + y₁ - m x₁ ) / √ ( m² + 1 )  + ( m - 1 + y₁ - m x₁ ) / √ ( m² + 1 )  = 0

= > 2 m + 3 ( y₁ - m x₁ ) - 2 + m - 1 = 0

Now using ( ii ) we get :

= > 3 ( y - m x ) + 3 m - 3 = 0

= > y - m x + m - 1 = 0

= > y - 1 = m ( x - 1 )  [ Comparing with ( i ) we get : ]

= > x₁ = 1 and y₁ = 1

Therefore , the required point P ( 1 , 1 ).