AP EAMCET201922 Apr 2019Morning ShiftMathematicsPair of LinesActual
A pair of lines (S= 0 ) together with the lines given by the equation (8 x^2-14 x y+3 y^2+10 x+10 y-25=0 ) form a parallelogram. If its diagonals intersect at the point ((3,2) ), then the equation (S=0 ), is
Options
- A(6 x^2-9 x y+y^2-25 x+30 y+25=0 )
- B(8 x^2-14 x y+3 y^2-25 x+30 y+50=0 )
- C(8 x^2-14 x y+3 y^2-50 x+50 y+75=0 )
- D(6 x^2+14 x y-3 y^2-30 x+40 y-75=0 )
Correct answer
C. (8 x^2-14 x y+3 y^2-50 x+50 y+75=0 )
Step-by-step solution
Equation of given pair of straight lines is ( array rlrl & 8 x^2-14 x y+3 y^2+10 x+10 y-25 & =0 & & (4 x-y-5)(2 x-3 y+5) & =0 array ) Now point of intersection of lines ( aligned 4 x-y-5 & =0 2 x-3 y+5 & =0 is (2,3) aligned ) So equation of (S=0 ) is ( (4 x-y+c₁ ) (2 x-3 y+c₂ )=0 ) and (S=0 ) passes through a point (P (x₁, y₁ ) ) such that mid-point of (P (x₁, y₁ ) ) and ((2,3) ) is ((3,2) ). so, ( array ll so, & x₁=4 and y₁=1 & c₁=-15 and c₂=-5 array ) So, required equation is ( array r (4 x-y-15)(2 x-3 y-5)=0 8 x