AP EAMCET · Maths · Pair of Lines
A pair of lines \(S=\mathbf{0}\) 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
- 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 \(6 x^2+14 x y-3 y^2-30 x+40 y-75=0\)
Answer & Solution
Correct Answer
(C) \(8 x^2-14 x y+3 y^2-50 x+50 y+75=0\)
Step-by-step Solution
Detailed explanation
Equation of given pair of straight lines is \(\begin{array}{rlrl} & 8 x^2-14 x y+3 y^2+10 x+10 y-25 & =0 \\ \Rightarrow & & (4 x-y-5)(2 x-3 y+5) & =0 \end{array}\) Now point of intersection of lines…
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