AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

The given equation of lines are,
3x2+2hxy−3y2=0
And
3x2+2hxy−3y2+2x−4y+c=0
The point of intersection of the lines given by
ax2+2hxy+by2+2gx+2fy+c=0‌ is ‌(

hf−bg

ab−h2

,

gh−af

ab−h2

)
Thus, the point of intersection of the second line is
A(x,y)=A(

2h−3

9+h2

,−

h+6

9+h2

)
The equation of diagonal of square passes through origin is,
y=

h+6

3−2h

x
The equation of diagonal of square that does passes through the origin is
2x−4y+c=0
The diagonal are perpendicular if,

1

2

(

h+6

3−2h

)=−1
h+6‌‌=4h−6
3h‌‌=12
h‌‌=4
Thus,
Point‌A‌‌=(

8−3

9+16

,−

10

9+16

)
‌‌=(

1

5

,−

2

5

)
Consider the figure shown below.

The midpoint of line OA is,
M=(

1

10

,−

1

5

)
Therefore,

2

10

+

4

5

+c=0
c=−1
‌ Therefore, ‌(h,c)=(4,−1)