Let the three sides of a triangle lie 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
- 20
- 20
- 5
- 5
Answer
(3) 5
Solution
Slopes: 4x - 7y + 10 = 0 m1 = 47; 7x + 4y = 15 m2 = - 74
m1 m2 = -1, so these two sides are perpendicular: the triangle is right-angled and its orthocentre is the vertex of the right angle.
Solving 4x - 7y = -10 and 7x + 4y = 15:
16x - 28y = -40 and 49x + 28y = 105 65x = 65
x = 1, y = 2; orthocentre H1 = (1, 2)
Triangle formed by x = 0, y = 0, x + y = 1 is right-angled at the origin, so H2 = (0, 0).
H1H2 = 12 + 22 = 5
Answer: (3)