. 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