JEE Main 29 June 2022 Shift 2 Solved Paper

For point A,
2x−y−1=0
x−2y+1=0
Solving (1) and (2), we get
x=1,y=1
∴ Point A=(1,1)
Altitude from B to line AC is perpendicular to line AC.
∴ Equator of altitude BH is
2x+y+λ=0
It passes through point H(‌

7

3

,‌

7

3

) so it satisfy the equation (3).
‌

14

3

+‌

7

3

+λ=0
⇒α=−7
∴ Altitude BH=2x+y−7=0
Solving equation (1) and (4), we get
x=2,y=3
∴ Point B=(2,3)
Altitude from C to line AB is perpendicular to line AB.
∴ Equation of altitude CH is
x+2y+λ=0
It passes through point H(‌

7

3

,‌

7

3

) so it satisfy equation (5).
‌

7

3

+‌

14

3

+λ=0
⇒λ=−7
∴ Altitude CH=x+2y−7=0
Solving equation (2) and (6), we get
x=3,y=2
∴ 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

3

=2,y=‌

1+2+3

3

=2
Now dDistance of point G(2,2) from center O(0,0) is
OG=√22+22=2√2