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

Concept: The triangle formed by joining the midpoints of the sides of a triangle is called the medial triangle. The centroid of the original triangle and the medial triangle coincide. If the vertices of the medial triangle are known, we can first find the vertices of the original triangle and then determine its incentre using the side–length weighted formula.
Step 1:Let the given midpoints be} \[ D\left(\frac{5}{2},7\right), \quad E\left(\frac{5}{2},3\right), \quad F(4,5) \] These correspond to midpoints of \(BC, CA,\) and \(AB\).
Step 2:Recover the vertices of triangle \(ABC\).} Using midpoint relations, \[ A = E + F - D \] \[ B = F + D - E \] \[ C = D + E - F \] Thus, \[ A(4,1), \quad B(4,9), \quad C(1,5) \]
Step 3:Compute side lengths.} \[ BC=\sqrt{(4-1)^2+(9-5)^2}=5 \] \[ CA=\sqrt{(4-1)^2+(1-5)^2}=5 \] \[ AB=\sqrt{(4-4)^2+(9-1)^2}=8 \]
Step 4:Use the incentre formula.} For triangle vertices \(A,B,C\), \[ I=\left( \frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c} \right) \] where \(a=BC,\, b=CA,\, c=AB\). Substituting, \[ I=\left(\frac{5(4)+5(4)+8(1)}{18}, \frac{5(1)+5(9)+8(5)}{18}\right) \] \[ I=(3,4) \] Thus, \[ h=3,\quad k=4 \]
Step 5:Compute the required value.} \[ 3h+k=3(3)+4=13 \]