If the orthocentre of the triangle, whose vertices are $(1,2),(2,3)$ and $(3,1)$ is $(\alpha, \beta)$, then the quadratic equation whose roots are $\alpha+4 \beta$ and $4 \alpha+\beta$, is

(A) $\mathrm{x}^{2}-19 \mathrm{x}+90=0$

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Straight Lines

If the orthocentre of the triangle, whose vertices are

(1, 2), (2, 3)

and

(3, 1)

(α, β)

, then the quadratic equation whose roots are

α + 4 β

and

4 α + β

, is

(A) $x^{2} - 19 x + 90 = 0$
(B) $x^{2} - 18 x + 80 = 0$
(C) $x^{2} - 22 x + 120 = 0$
(D) $x^{2} - 20 x + 99 = 0$ ✓ Correct

Answer: (D)

Solution

The altitude from

B

is perpendicular to

A C

(slope

- \frac{1}{2}

), so it has slope

2

and gives

β = 2 α - 1

; the altitude from

A

is perpendicular to

B C

(slope

- 2

), so it has slope

\frac{1}{2}

and gives

2 β - 4 = α - 1

. Solving these,

α = \frac{5}{3}

and

β = \frac{7}{3}

, so the roots are

α + 4 β = 11

and

4 α + β = 9

. With sum

20

and product

99

, the quadratic is

x^{2} - 20 x + 99 = 0

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option

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