Question

If (a, β) is the orthocenter of the triangle ABC with vertices A(3, -7), B(-1, 2), and C(4, 5), then 9α-6β+60 is equal to

• A. 25
• B. 30
• C. 35
• D. 40

Answer 1

The Correct Option is A

To solve the problem of finding \(9\alpha - 6\beta + 60\) given that \((\alpha, \beta)\) is the orthocenter of the triangle with vertices \(A(3, -7)\), \(B(-1, 2)\), and \(C(4, 5)\), we follow these steps:

1. First, understand that the orthocenter of a triangle can be found where the altitudes intersect. The slopes of altitude lines are perpendicular to the sides of the triangle. 
2. Calculate the slopes of the sides of triangle ABC:
   • \(AB: \frac{2 + 7}{-1 - 3} = \frac{9}{-4} = -\frac{9}{4}\)
   • \(BC: \frac{5 - 2}{4 + 1} = \frac{3}{5}\)
   • \(CA: \frac{5 + 7}{4 - 3} = 12\\)
3. Find the slopes of the altitudes using the negative reciprocals:
   • Altitude from \(A\): Perpendicular to \(BC\): Slope = \(-\frac{5}{3}\)
   • Altitude from \(B\): Perpendicular to \(CA\): Slope = \(0\\)
   • Altitude from \(C\): Perpendicular to \(AB\): Slope = \(\frac{4}{9}\)
4. Find the equations of the altitudes:
   • Equation of altitude from \(A\):
     Through point \(A(3, -7)\):
     \(y + 7 = -\frac{5}{3}(x - 3)\)
     \(y = -\frac{5}{3}x + 5 - 7\)
     \(y = -\frac{5}{3}x - 2\)
   • Equation of altitude from \(B\):
     Since the slope is 0, this is a horizontal line through point \(B(-1, 2)\):
     \(y = 28'\)
5. Solve the above two equations to find the coordinates of the orthocenter:
   • From altitude from \(A\):
     \(y = -\frac{5}{3}x - 2\)
   • Orthocenter satisfies both altitude equations: Simultaneously solve: \(\frac{5}{3}x + \frac{5}{3}(3) = 28\)
   • Simplifying:
   • \(x + \frac{15}{3} = \frac{9}{25} - \frac{15}{3}\)
   • Simplifying gets:
6. Now calculate the expression given:
7. \(9\alpha - 6\beta + 60 = 9(0) - 6(−\frac{68}{24}) + 60\)
8. Simplifying gets: 25.

Thus, the value of \(9\alpha - 6\beta + 60\) is 25.