Question

let mid "“ point of sides of $\Delta$ are $(\frac{5}{2}, 3), (\frac{5}{2}, 7) \, \& \, (4, 5)$. If incentre is $(h, k)$ then value of $3h + k$ is:

Answer 1

Step 1: Find the vertices of the triangle.
Let vertices be $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$.
Midpoints: $M_1 = (2.5, 3), M_2 = (2.5, 7), M_3 = (4, 5)$.
Using properties of midpoints: $x_1 = M_{x2} + M_{x3} - M_{x1} = 2.5 + 4 - 2.5 = 4$.
$y_1 = M_{y2} + M_{y3} - M_{y1} = 7 + 5 - 3 = 9$. So $A(4, 9)$.
Similarly, $B = (1, 5)$ and $C = (4, 1)$. (Note: Vertices calculation depends on which midpoint is which side).
Let's re-verify: $A(1, 5), B(4, 1), C(4, 9)$.
Check midpoints: $(A+B)/2 = (2.5, 3)$, $(B+C)/2 = (4, 5)$, $(A+C)/2 = (2.5, 7)$. Matches.


Step 2: Identify the type of triangle.
$A(1, 5), B(4, 1), C(4, 9)$.
Notice $BC$ is vertical ($x=4$). Length $BC = 9-1 = 8$.
The line $x=4$ is perpendicular to a horizontal line from $A(1, 5)$ to $(4, 5)$.
Actually, this is an isosceles triangle because $A$ lies on the perpendicular bisector of $BC$ ($y=5$).
Side $AB = \sqrt{(4-1)^2 + (1-5)^2} = \sqrt{3^2 + 4^2} = 5$.
Side $AC = \sqrt{(4-1)^2 + (9-5)^2} = \sqrt{3^2 + 4^2} = 5$.
Side $BC = 8$.


Step 3: Calculate the Incentre $(h, k)$.
$h = \frac{ax_1 + bx_2 + cx_3}{a+b+c} = \frac{8(1) + 5(4) + 5(4)}{8+5+5} = \frac{8 + 20 + 20}{18} = \frac{48}{18} = \frac{8}{3}$.
$k = \frac{ay_1 + by_2 + cy_3}{a+b+c} = \frac{8(5) + 5(1) + 5(9)}{18} = \frac{40 + 5 + 45}{18} = \frac{90}{18} = 5$.


Step 4: Find $3h + k$.
$3h + k = 3(\frac{8}{3}) + 5 = 8 + 5 = 13$.
The answer is 13.