Question:
The joint equation of two lines passing through \((-2,3)\) and parallel to the bisectors of the angle between the co-ordinate axes is
The joint equation of two lines passing through \((-2,3)\) and parallel to the bisectors of the angle between the co-ordinate axes is
Show Hint
The bisectors of the angle between the coordinate axes have slopes 1 and -1; write the two lines through the point and multiply.
Updated On: Nov 13, 2026
- \(x^{2}-y^{2}+4x+6y-4=0\)
- \(x^{2}+y^{2}+4x+6y-5=0\)
- \(x^{2}-y^{2}+4x+6y-5=0\)
- \(x^{2}+y^{2}+4x+6y+4=0\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The bisectors of the angle between the axes are \(y=x\) and \(y=-x\), with slopes \(1\) and \(-1\).
Step 2: Equations of the lines
Through \((-2,3)\): \(y-3=x+2\), i.e. \(x-y+5=0\).
And \(y-3=-(x+2)\), i.e. \(x+y-1=0\).
Step 3: Joint equation
\((x-y+5)(x+y-1)=0\)
\(x^2-y^2+4x+6y-5=0\).
Final Answer:
The joint equation is \(x^2-y^2+4x+6y-5=0\). \[ \boxed{\text{(C) }x^{2}-y^{2}+4x+6y-5=0} \]
The bisectors of the angle between the axes are \(y=x\) and \(y=-x\), with slopes \(1\) and \(-1\).
Step 2: Equations of the lines
Through \((-2,3)\): \(y-3=x+2\), i.e. \(x-y+5=0\).
And \(y-3=-(x+2)\), i.e. \(x+y-1=0\).
Step 3: Joint equation
\((x-y+5)(x+y-1)=0\)
\(x^2-y^2+4x+6y-5=0\).
Final Answer:
The joint equation is \(x^2-y^2+4x+6y-5=0\). \[ \boxed{\text{(C) }x^{2}-y^{2}+4x+6y-5=0} \]
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