MHT CET · Maths · Pair of Lines
The joint equation of pair of lines passing through (2,3) and parallel to the line
=0
- A \(x^{2}-y^{2}-4 x+6 y-5=0\)
- B \(x^{2}-y^{2}-4 x+6 y=0\)
- C \(x^{2}-y^{2}-4 x+6 y+17=0\)
- D \(x^{2}-y^{2}-4 x+6 y+2=0\)
Answer & Solution
Correct Answer
(A) \(x^{2}-y^{2}-4 x+6 y-5=0\)
Step-by-step Solution
Detailed explanation
\(x^{2}-y^{2}=0 \Rightarrow(x-y)(x+y)=0\)
Thus slopes of given lines are 1 and \(-1\).
The eq. of lines passing through \((2,3)\) and parallel to given liens are
\((y-3)=(x-2)\) and \((y-3)=-(x-2)\) i.e.
\(x-y+1=0\) and \((x+y-5)=0\)
Hence required eq. is
\((x-y+1)(x+y-5)=0\)
\(x^{2}-x y+x+x y-y^{2}+y-5 x+5 y-5=0\)
\(x^{2}-y^{2}-4 x+6 y-5=0\)
Thus slopes of given lines are 1 and \(-1\).
The eq. of lines passing through \((2,3)\) and parallel to given liens are
\((y-3)=(x-2)\) and \((y-3)=-(x-2)\) i.e.
\(x-y+1=0\) and \((x+y-5)=0\)
Hence required eq. is
\((x-y+1)(x+y-5)=0\)
\(x^{2}-x y+x+x y-y^{2}+y-5 x+5 y-5=0\)
\(x^{2}-y^{2}-4 x+6 y-5=0\)
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