AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

. Simplify the given equation of pair of lines.
8x2−14xy+3y2+10x +10y−25=0
(4x−y−5)(2x−3y+5)=0
Now, solving the two equations, 4x−y−5=0 and 2x−3y+5=0, we get the point of intersection as (2,3)
So, equation of S=0 will be,
(4x−y+c1)(2x−3y+c2)=0
As the lines S=0 passes through point P(x1,y1) such that mid point of P (x1,y1) and (2,3) is (3,2) therefore,
x1=4 and y1=1
This gives,
c1=−15 and c2=−5
So required equation of pair of lines is,
(4x−y−15)(2x−3y−5)=0
8x2−14xy+3y2−50x +50y+75=0