JEE Main 2026 Apr 2 Shift 1, Mathematics Q9: Solution of Triangles

JEE Main2026Apr 2, Shift 1Mathematics

Let the mid points of the sides of a triangle $A B C$ be $(\frac{5}{2}, 7)$ , $(\frac{5}{2}, 3)$ and $(4, 5)$ . If its incentre is $(h, k)$ , then $3 h + k$ is equal to :

A
$11$
B
$12$
C
$13$
D
$14$

Solution

If $M_{1}, M_{2}, M_{3}$ are midpoints of $B C, C A, A B$ , then each vertex is obtained as the sum of two midpoints minus the third. With $M_{1} = (5/2, 7)$ , $M_{2} = (5/2, 3)$ , $M_{3} = (4, 5)$ :

$A = M_{2} + M_{3} - M_{1} = (4, 1)$ is one possibility; rearranging gives vertices $A (1, 5)$ , $B (4, 1)$ , $C (4, 9)$ .

Side lengths: $a = B C = 8$ , $b = C A = 9 + 16 = 5$ , $c = A B = 9 + 16 = 5$ .

Incentre $= \frac{a A + b B + c C}{a + b + c} = \frac{8 ( 1 , 5 ) + 5 ( 4 , 1 ) + 5 ( 4 , 9 )}{18} = \frac{( 8 + 20 + 20 , 40 + 5 + 45 )}{18} = (\frac{48}{18}, \frac{90}{18}) = (\frac{24}{9}, 5)$ .

$∴ 3 h + k = 3 \cdot \frac{24}{9} + 5 = 8 + 5 = 13$ .

Concept behind this question

Solution of TrianglesNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Apr 2 Shift 1, Mathematics Q9: Solution of Triangles

Concept behind this question

More previous year questions on Solution of Triangles

Practice more Mathematics