A right-angled triangle has vertices at (2, 3), (9, −1) and (5, k).
Find the sum of all the possible values of k.
- A−8
- B−6
- C0.25
- D2
- E2.25
- F8.25
- G10.25
Show the answer and worked solution
answer · E
- A−8
- B−6
- C0.25
- D2
- E2.25
- F8.25
- G10.25
Call the points A(2, 3), B(9, −1) and C(5, k). The right angle may be at any vertex. At A: AB⋅AC = 7(3) + (−4)(k−3) = 0 gives k = 334 = 8.25. At B: BA⋅BC = (−7)(−4) + 4(k+1) = 0 gives k = −8. At C: CA⋅CB = (−3)(4) + (3−k)(−1−k) = k2 − 2k − 15 = 0 gives k = 5 or k = −3. The sum is 8.25 − 8 + 5 − 3 = 2.25.