AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

The vertices of the triangle are A(2,3),B(2,4),C(4,3) The length of the sides
AB=1,BC=√5,CA=2
coordinate of the orthocentre is (

x1+x2+x3

3

,

y1+y2+y3

3

) =(

8

3

,

10

3

)
It is given that the centroid of the triangle divides the linejoining point O and S in ration 2: 1 .

8

3

=

2x+2

3

⇒x=3

10

3

=

2y+3

3

⇒y=

7

2

The point S(3,

7

2

).
Therefore, the value of1=AO2+9BG2+4CS2
=0+9×[(2−

8

3

)2 +(4−

10

3

)2]+4[(4−3)2+(3−

7

2

)2]
=13