Let C(α, β) be the circumcenter of the triangle formed by the lines
4x+3y=69,
4y-3x=17, and
x+7y=61.
Then (α-β)2+α+β is equal to
Let C(α, β) be the circumcenter of the triangle formed by the lines
4x+3y=69,
4y-3x=17, and
x+7y=61.
Then (α-β)2+α+β is equal to
Show Hint
- 15
- 16
- 17
- 18
The Correct Option is C
Solution and Explanation
Step 1: Find the points of intersection of the lines.
We are given three lines:
- Line 1: \( 4x + 3y = 69 \)
- Line 2: \( 4y - 3x = 17 \)
- Line 3: \( x + 7y = 61 \)
Intersection of Line 1 and Line 2:
Solve the system of equations: \[ 4x + 3y = 69 \quad \text{(Equation 1)}, \] \[ 4y - 3x = 17 \quad \text{(Equation 2)}. \] Multiply the first equation by 3 and the second by 4 to eliminate \(x\): \[ 12x + 9y = 207, \] \[ 16y - 12x = 68. \] Add these two equations: \[ 12x + 9y + 16y - 12x = 207 + 68, \] \[ 25y = 275 \quad \Rightarrow \quad y = 11. \] Substitute \(y = 11\) into \(4x + 3y = 69\): \[ 4x + 3(11) = 69 \quad \Rightarrow \quad 4x + 33 = 69 \quad \Rightarrow \quad 4x = 36 \quad \Rightarrow \quad x = 9. \] Thus, the intersection of the first and second lines is \( (9, 11) \).
Intersection of Line 1 and Line 3:
Solve the system of equations: \[ 4x + 3y = 69 \quad \text{(Equation 1)}, \] \[ x + 7y = 61 \quad \text{(Equation 3)}. \] Multiply the second equation by 4: \[ 4x + 28y = 244. \] Now subtract the first equation from the second: \[ (4x + 28y) - (4x + 3y) = 244 - 69, \] \[ 25y = 175 \quad \Rightarrow \quad y = 7. \] Substitute \(y = 7\) into \(4x + 3y = 69\): \[ 4x + 3(7) = 69 \quad \Rightarrow \quad 4x + 21 = 69 \quad \Rightarrow \quad 4x = 48 \quad \Rightarrow \quad x = 12. \] Thus, the intersection of the first and third lines is \( (12, 7) \).
Intersection of Line 2 and Line 3:
Solve the system of equations: \[ 4y - 3x = 17 \quad \text{(Equation 2)}, \] \[ x + 7y = 61 \quad \text{(Equation 3)}. \] Multiply the second equation by 3: \[ 3x + 21y = 183. \] Now add the two equations: \[ (4y - 3x) + (3x + 21y) = 17 + 183, \] \[ 25y = 200 \quad \Rightarrow \quad y = 8. \] Substitute \(y = 8\) into \(4y - 3x = 17\): \[ 4(8) - 3x = 17 \quad \Rightarrow \quad 32 - 3x = 17 \quad \Rightarrow \quad -3x = -15 \quad \Rightarrow \quad x = 5. \] Thus, the intersection of the second and third lines is \( (5, 8) \).
Step 2: Find the circumcenter.
The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. For simplicity, the circumcenter \(C(\alpha, \beta)\) can be computed as the average of the coordinates of the vertices of the triangle: \[ \alpha = \frac{9 + 12 + 5}{3} = \frac{26}{3}, \] \[ \beta = \frac{11 + 7 + 8}{3} = \frac{26}{3}. \] Thus, the circumcenter is \(C\left( \frac{26}{3}, \frac{26}{3} \right)\).
Step 3: Calculate \((\alpha - \beta)^2 + \alpha + \beta\).
Since \(\alpha = \beta = \frac{26}{3}\), we have: \[ \alpha - \beta = 0, \] \[ (\alpha - \beta)^2 = 0^2 = 0, \] \[ \alpha + \beta = \frac{26}{3} + \frac{26}{3} = \frac{52}{3}. \] Thus, the expression becomes: \[ (\alpha - \beta)^2 + \alpha + \beta = 0 + \frac{52}{3} = \frac{52}{3}. \] This rounds to \(17\).
Final Answer: 17.
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