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Step-by-Step Solution
Step 1: Determine the slopes of the sides of the given triangle
We have triangle ABC with vertices
A(3, –7), B(–1, 2), and C(4, 5). First, find the slopes of sides AB, BC, and CA.
Slope of AB:
$m_{AB} = \frac{2 - (-7)}{-1 - 3} = \frac{9}{-4} = -\frac{9}{4}$
Slope of BC:
$m_{BC} = \frac{5 - 2}{4 - (-1)} = \frac{3}{5}$
Slope of CA:
$m_{CA} = \frac{-7 - 5}{3 - 4} = \frac{-12}{-1} = 12$
Step 2: Form the equations of the altitudes
An altitude from a vertex is perpendicular to the opposite side. The slope of a line perpendicular to a line of slope $m$ is
$-\frac{1}{m}$. We will form equations of two altitudes and solve for their intersection (the orthocenter).
Altitude from vertex C (perpendicular to AB)
Slope of AB is $-\frac{9}{4}$. A line perpendicular to this has slope
$m_{altitude\_C} = \frac{4}{9}$. Using point-slope form from point C(4, 5):
$y - 5 = \frac{4}{9}(x - 4).$
Multiply through by 9 to clear fractions:
$9y - 45 = 4x - 16.$
Rearrange to standard form:
$4x - 9y + 29 = 0.\quad (1)
Altitude from vertex A (perpendicular to BC)
Slope of BC is $\frac{3}{5}$. A line perpendicular to this has slope
$m_{altitude\_A} = -\frac{5}{3}$. Using point-slope form from point A(3, –7):
$y + 7 = -\frac{5}{3}(x - 3).$
Multiply through by 3:
$3y + 21 = -5x + 15.$
Rearrange to standard form:
$5x + 3y + 6 = 0.\quad (2)
Step 3: Find the orthocenter $(\alpha,\beta)$ from the intersection of the altitudes
The two altitude equations are:
$4x - 9y + 29 = 0.$
$5x + 3y + 6 = 0.$
From equation (2), express $y$ in terms of $x$:
$5x + 3y + 6 = 0 \implies 3y = -5x - 6 \implies y = -\frac{5}{3}x - 2.
Substitute this into equation (1):
$4x - 9\left(-\frac{5}{3}x - 2\right) + 29 = 0.
Simplify step by step:
$4x + 15x + 18 + 29 = 0 \implies 19x + 47 = 0 \implies x = -\frac{47}{19}.
Now substitute $x = -\frac{47}{19}$ back into $y = -\frac{5}{3}x - 2$:
$y = -\frac{5}{3} \left(-\frac{47}{19}\right) - 2
= \frac{235}{57} - 2
= \frac{235 - 114}{57}
= \frac{121}{57}.
Hence, the orthocenter is
$(\alpha,\beta) = \left(-\frac{47}{19}, \frac{121}{57}\right).$
Step 4: Calculate $9\alpha - 6\beta + 60$
Substitute $\alpha = -\frac{47}{19}$ and $\beta = \frac{121}{57}$:
$9\alpha - 6\beta + 60
= 9\left(-\frac{47}{19}\right)
- 6\left(\frac{121}{57}\right)
+ 60.
Compute each term individually:
$9\left(-\frac{47}{19}\right) = -\frac{423}{19}$.
$-6\left(\frac{121}{57}\right) = -\frac{726}{57}$.
$60 = \frac{60 \times 57}{57} = \frac{3420}{57}$.
Next, convert $-\frac{423}{19}$ to a denominator of 57:
$-\frac{423}{19} = -\frac{423 \times 3}{19 \times 3}
= -\frac{1269}{57}.$
Now combine all three terms with denominator 57:
$-\frac{1269}{57} - \frac{726}{57} + \frac{3420}{57}
= \frac{-1269 - 726 + 3420}{57}
= \frac{3420 - 1995}{57}
= \frac{1425}{57}
= 25.
Final Answer
$9\alpha - 6\beta + 60 = 25.$
