Let $(\alpha, \beta)$ be the centroid of the triangle formed by the lines $15x-y=82$, $6x-5y=-4$ and $9x+4y=17$. Then $\alpha+2\beta$ and $2\alpha-\beta$ are the roots of the equation

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Straight Lines

Let

(α, β)

be the centroid of the triangle formed by the lines

15 x - y = 82

6 x - 5 y = - 4

and

9 x + 4 y = 17

. Then

α + 2 β

and

2 α - β

are the roots of the equation

(A) $x^{2} - 7 x + 12 = 0$
(B) $x^{2} - 13 x + 42 = 0$ ✓ Correct
(C) $x^{2} - 14 x + 48 = 0$
(D) $x^{2} - 10 x + 25 = 0$

Answer: (B)

Solution

Solving the three lines pairwise, we get the vertices of the triangle:

(6, 8)

(1, 2)

, and

(5, - 7)

.

The centroid is:

α = \frac{6 + 1 + 5}{3} = 4, β = \frac{8 + 2 - 7}{3} = 1

Therefore:

α + 2 β = 4 + 2 = 6

2 α - β = 8 - 1 = 7

The quadratic with roots 6 and 7 is:

x^{2} - (6 + 7) x + (6 \cdot 7) = 0 \Rightarrow x^{2} - 13 x + 42 = 0

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Single correct option

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