Question

Let the midpoints of the sides of a triangle \(ABC\) be \( \left(\frac{5}{2},7\right), \left(\frac{5}{2},3\right)\) and \( (4,5) \). If its incentre is \((h,k)\), then \(3h+k\) is equal to:

• A. \(11\)
• B. \(12\)
• C. \(13\)
• D. \(14\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Given mid-points of sides, we can find the vertices of the triangle. Then, using side lengths and vertex coordinates, we apply the Incentre formula: \(I = \left( \frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c} \right)\).
Step 2: Key Formula or Approach:
1. Vertices: \(x_1+x_2=2x_{m1}\), etc. Or \(A = M_2+M_3-M_1\).
2. Midpoints: \(M_1(2.5, 7), M_2(2.5, 3), M_3(4, 5)\).
Step 3: Detailed Explanation:
Let the vertices be \(A, B, C\).
\(A = M_2 + M_3 - M_1 = (2.5+4-2.5, 3+5-7) = (4, 1)\).
\(B = M_1 + M_3 - M_2 = (2.5+4-2.5, 7+5-3) = (4, 9)\).
\(C = M_1 + M_2 - M_3 = (2.5+2.5-4, 7+3-5) = (1, 5)\).
Vertices are \(A(4, 1), B(4, 9), C(1, 5)\).
Side lengths:
\(c = AB = \sqrt{(4-4)^2 + (9-1)^2} = 8\).
\(a = BC = \sqrt{(4-1)^2 + (9-5)^2} = \sqrt{3^2 + 4^2} = 5\).
\(b = AC = \sqrt{(4-1)^2 + (1-5)^2} = \sqrt{3^2 + 4^2} = 5\).
Incentre \((h, k)\):
\(h = \frac{5(4) + 5(4) + 8(1)}{5+5+8} = \frac{20+20+8}{18} = \frac{48}{18} = \frac{8}{3}\).
\(k = \frac{5(1) + 5(9) + 8(5)}{18} = \frac{5+45+40}{18} = \frac{90}{18} = 5\).
Calculate \(3h + k = 3(8/3) + 5 = 8 + 5 = 13\).
Step 4: Final Answer:
The value of \(3h + k\) is 13.