Let the mid points of the sides of a triangle ABC be (5/2, 7), (5/2, 3) and (4, 5). If its incentre is (h, k), then 3h+k is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q9 – JEE-MAIN 2026

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Home Maths Straight LinesQuestion #19

Let the mid points of the sides of a triangle ABC 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 :

Clue 1 of 3: Midpoint Theorem for Triangles

The sum of the coordinates of the vertices of a triangle is twice the sum of the coordinates of the midpoints of its sides.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine the Vertices of the Triangle✦ Active

Let the vertices be $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ , $C (x_{3}, y_{3})$ and the midpoints be $D (\frac{5}{2}, 7)$ , $E (\frac{5}{2}, 3)$ , $F (4, 5)$ . Using the midpoint formula, we have:

x_{1} + x_{2} = 5, x_{2} + x_{3} = 5, x_{3} + x_{1} = 8

Summing these gives $2 (x_{1} + x_{2} + x_{3}) = 18 ⟹ x_{1} + x_{2} + x_{3} = 9$ . Solving for $x_{i}$ : $x_{1} = 9 - 5 = 4$ , $x_{2} = 9 - 8 = 1$ , $x_{3} = 9 - 5 = 4$ . Similarly for y-coordinates:

y_{1} + y_{2} = 14, y_{2} + y_{3} = 6, y_{3} + y_{1} = 10

Summing these gives $2 (y_{1} + y_{2} + y_{3}) = 30 ⟹ y_{1} + y_{2} + y_{3} = 15$ . Solving for $y_{i}$ : $y_{1} = 15 - 6 = 9$ , $y_{2} = 15 - 10 = 5$ , $y_{3} = 15 - 14 = 1$ . Thus, the vertices are $A (4, 9)$ , $B (1, 5)$ , and $C (4, 1)$ .

Step 2: Calculate Side Lengths and Incenter Coordinates○ Expand

Calculate the side lengths using the distance formula:

a = B C = \sqrt{(4 - 1)^{2} + (1 - 5)^{2}} = \sqrt{3^{2} + (- 4)^{2}} = \sqrt{9 + 16} = 5

b = A C = \sqrt{(4 - 4)^{2} + (1 - 9)^{2}} = \sqrt{0^{2} + (- 8)^{2}} = \sqrt{64} = 8

c = A B = \sqrt{(1 - 4)^{2} + (5 - 9)^{2}} = \sqrt{(- 3)^{2} + (- 4)^{2}} = \sqrt{9 + 16} = 5

The incenter $(h, k)$ is given by the formula:

h = \frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c} = \frac{5 (4) + 8 (1) + 5 (4)}{5 + 8 + 5} = \frac{20 + 8 + 20}{18} = \frac{48}{18} = \frac{8}{3}

k = \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c} = \frac{5 (9) + 8 (5) + 5 (1)}{5 + 8 + 5} = \frac{45 + 40 + 5}{18} = \frac{90}{18} = 5

So, the incenter is $(h, k) = (\frac{8}{3}, 5)$ .

Step 3: Compute the Final Expression○ Expand

The required value is $3 h + k$ :

3 h + k = 3 (\frac{8}{3}) + 5 = 8 + 5 = 13

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Maths Question 9 – JEE-MAIN 2026

Clue 1 of 3: Midpoint Theorem for Triangles

🧠 Full Solution Path