Let the mid points of the sides of a triangle ABC be , and (4, 5). If its incentre is (h, k), then 3h + k is equal to: [2026]

Question

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 3h + k is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

Let coordinates of A, B and C are $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ , respectively.

FIGURE

$x_{1} + x_{2} = 5$ ... (i)

$x_{1} + x_{3} = 5$ ... (ii)

$x_{2} + x_{3} = 8$ ... (iii)

Adding (i), (ii) and (iii), we get

$x_{1} + x_{2} + x_{3} = 9$

On solving, we get

$x_{1} = 1, x_{2} = 4, x_{3} = 4$

Similarly,

$y_{1} + y_{2} = 6$ ... (iv)

$y_{1} + y_{3} = 14$ ...(v)

$y_{2} + y_{3} = 10$ ... (vi)

Adding (iv), (v) and (vi), we get

$y_{1} + y_{2} + y_{3} = 15$

On solving, we get

$y_{1} = 5, y_{2} = 1, y_{3} = 9$

Thus coordinates of A, B and C are (1, 5), (4, 1) and (4,9) respectively.

FIGURE

Incente (I) is

$(\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c})$

Where a = |BC|, b = |AC| and c = |AB|

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

$a = B C = \sqrt{{(4 - 4)}^{2} + {(9 - 1)}^{2}} = \sqrt{0 + 64} = 8$

$b = A C = \sqrt{{(4 - 1)}^{2} + {(9 - 5)}^{2}} = \sqrt{9 + 16} = 5$

$∴ I = (\frac{8 \times 1 + 5 \times 4 + 5 \times 4}{8 + 5 + 5}, \frac{8 \times 5 + 5 \times 1 + 5 \times 9}{8 + 5 + 5})$

$= (\frac{48}{18}, \frac{90}{18}) = (\frac{8}{5}, 5)$

$∵$ Incentre is (h, k)

$∴$ $h = \frac{8}{3} and k = 5$

$\Rightarrow 3 h + k = 3 \times \frac{8}{3} + 5 = 8 + 5 = 13$

Solution

Q.
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 3h + k is equal to: [2026]

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