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Solution


Q.

Let the three sides of a triangle are on the lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x = 0, y = 0 and x + y = 1 is          [2025]
1 5  
2 20  
3 5  
4 20  

Ans.

(1)

Since the lines 4x –7y + 10 = 0 and 7x + 4y = 15 are perpendicular to each other.

So, the triangle formed by lines 4x –7y + 10 = 0, 7x + 4y = 15 and x + y = 5 is right angled triangle.

So, orthrocentre = Point of intersection of 4x – 7y + 10 = 0 and 7x + 4y = 15 which is, B = (1, 2)

Also, triangle formed by sides x = 0, y = 0, x + y = 1 is right angled triangle.

So, orthrocentre, Q = (0, 0)

∴   Distance between Q and B=1+4=5 units..