QJEEM22Ma0b0a: JEE Main 2022 Mathematics - Class 11 JEEM22Ma0b0a Solution: The distance of the origin from the centroid of the triangle…

For point A,

2x $-$ y $-$ 1 = 0 ...... (1)

x $-$ 2y + 1 = 0 ...... (2)

Solving (1) and (2), we get

x = 1, y = 1.

$\therefore$ Point A = (1, 1)

Altitude from B to line AC is perpendicular to line AC.

$\therefore$ Equator of altitude BH is

2x + y + $\lambda$ = 0 ...... (3)

It passes through point $H\left( {{7 \over 3},{7 \over 3}} \right)$ so it satisfy the equation (3).

${{14} \over 3} + {7 \over 3} + \lambda = 0$

$\Rightarrow$ $\alpha$ = $-$ 7

$\therefore$ Altitude BH = 2x + y $-$ 7 = 0 ...... (4)

Solving equation (1) and (4), we get

x = 2, y = 3.

$\therefore$ Point B = (2, 3)

Altitude from C to line AB is perpendicular to line AB.

$\therefore$ Equation of altitude CH is

x + 2y + $\lambda$ = 0 ...... (5)

It passes through point $H\left( {{7 \over 3},{7 \over 3}} \right)$ so it satisfy equation (5).

${7 \over 3} + {{14} \over 3} + \lambda = 0$

$\Rightarrow$ $\lambda$ = $-$ 7

$\therefore$ Altitude CH = x + 2y $-$ 7 = 0 ...... (6)

Solving equation (2) and (6), we get

x = 3, y = 2

$\therefore$ Point C = (3, 2)

Centroid G (x, y) of triangle A (1, 1), B (2, 3) and C (3, 2) is

$x = {{1 + 2 + 3} \over 3} = 2$ , $y = {{1 + 2 + 3} \over 3} = 2$

Now, distance of point G (2, 2) from center O (0, 0) is

$OG = \sqrt {{2^2} + {2^2}} = 2\sqrt 2$