JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Straight Lines

The distance of the origin from the centroid of the triangle whose two sides have the equations

x - 2 y + 1 = 0

and

2 x - y - 1 = 0

and whose orthocenter is

(\frac{7}{3}, \frac{7}{3})

is:

Answer: (C)

Solution

The two sides of the triangle are

A B \equiv x - 2 y + 1 = 0

and

A C \equiv 2 x - y - 1 = 0

. Solving these gives vertex

A (1, 1)

.

The altitude from

B

passes through the orthocenter

H (\frac{7}{3}, \frac{7}{3})

and is perpendicular to

A C

. Since

A C

has slope 2, the altitude has slope

- \frac{1}{2}

, giving equation

B H : x + 2 y - 7 = 0

. Since

B

lies on both

A B

and

B H

, solving these gives

B (3, 2)

.

Similarly, the altitude from

C

is perpendicular to

A B

(slope

\frac{1}{2}

), so has slope

- 2

, giving

C H : 2 x + y - 7 = 0

. Solving with

A C

gives

C (2, 3)

.

The centroid is

E (\frac{1 + 3 + 2}{3}, \frac{1 + 2 + 3}{3}) = (2, 2)

.

The distance from the origin is

O E = 2^{2} + 2^{2} = 22

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option