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Triangle from Midpoints, Incentre: 3h + k = 13 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let the mid points of the sides of a triangle ABCABC be (52,7),(52,3)\left(\dfrac52,7\right),\left(\dfrac52,3\right) and (4,5)(4,5). If its incentre is (h,k)(h,k), then 3h+k3h+k is equal to
A1111
B1212
C1313correct
D1414
Solution
Step 1: Recover the vertices from the midpoints. If the midpoints of BC,CA,ABBC, CA, AB are given, each vertex is the sum of two adjacent midpoints minus the third. This yields
A(1,5),B(4,1),C(4,9).A(1,5),\quad B(4,1),\quad C(4,9).
 Step 2: Compute side lengths (opposite to each vertex):
a=BC=(4−4)2+(9−1)2=8,b=CA=5,c=AB=5.a=BC=\sqrt{(4-4)^2+(9-1)^2}=8,\quad b=CA=5,\quad c=AB=5.
 Step 3: Incentre =a A+b B+c Ca+b+c=\dfrac{a\,A+b\,B+c\,C}{a+b+c}:
h=8(1)+5(4)+5(4)18=4818=249,k=8(5)+5(1)+5(9)18=9018=5.h=\frac{8(1)+5(4)+5(4)}{18}=\frac{48}{18}=\frac{24}{9},\qquad k=\frac{8(5)+5(1)+5(9)}{18}=\frac{90}{18}=5.
 Step 4:
3h+k=3⋅249+5=8+5=13.3h+k=3\cdot\frac{24}{9}+5=8+5=13.
 Correct answer: (3)
Solution working
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