Let k be an integer such that the triangle with vertices $(\mathrm{k},-3 \mathrm{k}),(5, \mathrm{k})$ and $(-\mathrm{k}, 2)$ has area 28 sq. units. Then the orthocentre of this triangle is at the point :

(A) $\left(2, \frac{1}{2}\right)$

(B) $\left(2,-\frac{1}{2}\right)$

(C) $\left(1, \frac{3}{4}\right)$

(D) $\left(1,-\frac{3}{4}\right)$

JEE Main · 2017 · 02 Apr 2017 · Offline · Straight Lines

Let k be an integer such that the triangle with vertices

(k, - 3 k), (5, k)

and

(- k, 2)

has area 28 sq. units. Then the orthocentre of this triangle is at the point :

(A) $(2, \frac{1}{2})$ ✓ Correct
(B) $(2, - \frac{1}{2})$
(C) $(1, \frac{3}{4})$
(D) $(1, - \frac{3}{4})$

Answer: (A)

Solution

The area condition

\frac{1}{2} ∣5 k^{2} + 13 k + 10∣ = 28

becomes

5 k^{2} + 13 k - 46 = 0

, whose integer root is

k = 2

(the other root is not whole). The vertices are then

A (2, - 6)

B (5, 2)

C (- 2, 2)

. Since

B C

is horizontal, the altitude from

A

is the vertical line

x = 2

; the side

A C

has slope

- 2

, so the altitude from

B

has slope

\frac{1}{2}

through

(5, 2)

. At

x = 2

that altitude gives

y = \frac{1}{2}

, so the orthocentre is

(2, \frac{1}{2})

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option

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