Let the vertex A of a triangle ABC be (1, 2) and the midpoint of the side AB be(5, –1). If the centroid of this triangle is (3, 4) and its circumcenter is , then is equal to: [2026]

Question

Let the vertex A of a triangle ABC be (1, 2) and the midpoint of the side AB be(5, –1). If the centroid of this triangle is (3, 4) and its circumcenter is $(α, β)$ , then $21 (α + β)$ is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

From the given information:

FIGURE

Coordinates of $B : \frac{x_{2} + 1}{2} = 5; \frac{y_{2} + 2}{2} = - 1$

$\Rightarrow x_{2} = 9, y_{2} = - 4$

$∴ B \equiv (9, - 4)$

Coordinates of $C : \frac{1 + 9 + x_{3}}{3} = 3$

$\Rightarrow x_{3} = - 1 and \frac{2 - 4 + y_{3}}{3} = 4$

$\Rightarrow y_{3} = 14$

$∴ C \equiv (- 1, 14)$

Slope of $⊥^{r}$ bisector of AB, $m_{1} = \frac{4}{3}$

Equation of $⊥^{r}$ bisector of AB

$y + 1 = \frac{4}{3} (x - 5)$

$\Rightarrow 4 x - 3 y + 23 = 0$ ... (i)

Midpoint of BC is N (4, 5)

Slope of $⊥^{r}$ bisector of BC, $m_{2} = \frac{5}{9}$

Equation of $⊥^{r}$ bisector of BC,

$y - 5 = \frac{5}{9} (x - 4) \Rightarrow 9 y - 45 = 5 x - 20$

$\Rightarrow 5 x - 9 y + 25 = 0$ ... (ii)

Solving (i) & (ii), we get

$x = \frac{94}{7}, y = \frac{4 \times 94}{21} - \frac{23}{3} = \frac{215}{21}$

$∴ (α, β) = (\frac{94}{7}, \frac{215}{21})$

Thus, $21 (α + β) = 497$

Solution

Q.
Let the vertex A of a triangle ABC be (1, 2) and the midpoint of the side AB be(5, –1). If the centroid of this triangle is (3, 4) and its circumcenter is $(α, β)$ , then $21 (α + β)$ is equal to: [2026]

1 309

2 403

3 497

4 524

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