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Let the equations of two sides of a triangle be 3x – 2y + 6 = 0 and 4x + 5y – 20 = 0. If the orthocentre of this triangle is at (1, 1), then the equation of its third side is : 

(1) 122y – 26x – 1675 = 0 

(2) 26x – 122y – 1675 = 0 

(3) 122y + 26x + 1675 = 0 

(4) 26x + 61y + 1675 = 0

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Best answer

The correct option (2) 26x – 122y – 1675 = 0

Explanation:

Equations of AB and AC are : 

3x – 2y + 6 = 0 and 4x + 5y – 20 = 0, respectively.

Orthocenter (O) is (1, 1). Since BE ⊥ AC passing through O, BE : 5x – 4y – 1 = 0. 

Similarly, CF : 2x + 3y – 5 = 0 

Use family of lines concept 

⇒ BC : 26x – 122y = 1675

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