Let be an integer such that the triangle with vertices , and has area 28 sq units. Then, the orthocentre of this triangle is at the point
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
Using the area formula, (1/2)|det| = 28, so |det| = 56.
Setting up the determinant with vertices (k, -3k), (5, k), (-k, 2):
Area = (1/2)|k(k-2) + 5(2-(-3k)) + (-k)((-3k)-k)|
= (1/2)|k²-2k + 10+15k + (-k)(-4k)|
= (1/2)|k²-2k + 15k + 10 + 4k²|
= (1/2)|5k² + 13k + 10| = 28
So 5k² + 13k + 10 = ±56.
Case 1: 5k² + 13k - 46 = 0 → k = (-13 ± √(169+920))/10 = (-13 ± 33)/10 → k = 2 or k = -46/10 (not integer).
Case 2: 5k² + 13k + 66 = 0 → discriminant = 169 - 1320 < 0, no real solutions.
So k = 2, giving vertices A(2, -6), B(5, 2), C(-2, 2).
Since B and C have the same y-coordinate (y=2), BC is horizontal. The altitude from A to BC is vertical: x = 2.
Slope of AB = (2-(-6))/(5-2) = 8/3, so altitude from C(-2,2) perpendicular to AB has slope -3/8.
Altitude from C: y -
Setting up the determinant with vertices (k, -3k), (5, k), (-k, 2):
Area = (1/2)|k(k-2) + 5(2-(-3k)) + (-k)((-3k)-k)|
= (1/2)|k²-2k + 10+15k + (-k)(-4k)|
= (1/2)|k²-2k + 15k + 10 + 4k²|
= (1/2)|5k² + 13k + 10| = 28
So 5k² + 13k + 10 = ±56.
Case 1: 5k² + 13k - 46 = 0 → k = (-13 ± √(169+920))/10 = (-13 ± 33)/10 → k = 2 or k = -46/10 (not integer).
Case 2: 5k² + 13k + 66 = 0 → discriminant = 169 - 1320 < 0, no real solutions.
So k = 2, giving vertices A(2, -6), B(5, 2), C(-2, 2).
Since B and C have the same y-coordinate (y=2), BC is horizontal. The altitude from A to BC is vertical: x = 2.
Slope of AB = (2-(-6))/(5-2) = 8/3, so altitude from C(-2,2) perpendicular to AB has slope -3/8.
Altitude from C: y -
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