Question:
The joint equation of a pair of lines passing through point \((1,4)\), one of which is parallel to X-axis and the other makes an angle of \(45^{\circ}\) with the positive direction of X-axis, is
The joint equation of a pair of lines passing through point \((1,4)\), one of which is parallel to X-axis and the other makes an angle of \(45^{\circ}\) with the positive direction of X-axis, is
Show Hint
One line has slope 0, the other slope 1; multiply the two line equations.
Updated On: Oct 1, 2026
- \(x^2-xy-x+4y-12 = 0\)
- \(xy-y^2-4x+7y-12 = 0\)
- \(x^2+2xy-y^2+7 = 0\)
- \(xy-2y^2+3x+2y+17 = 0\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
The joint equation of two lines is the product of their equations set equal to zero.
Step 2: Key Formula or Approach
Line 1: parallel to X-axis through \((1,4)\), so \(y-4=0\). Line 2: slope \(\tan45^{\circ}=1\) through \((1,4)\), so \(y-4=x-1\), i.e. \(x-y+3=0\).
Step 3: Detailed Explanation
\[ (y-4)(x-y+3)=0 \]
\[ xy-y^2+3y-4x+4y-12=0 \]
\[ xy-y^2-4x+7y-12=0 \]
Final Answer:
The joint equation is \(xy-y^2-4x+7y-12=0\), option (B). \[ \boxed{xy-y^2-4x+7y-12=0\ \text{(B)}} \]
The joint equation of two lines is the product of their equations set equal to zero.
Step 2: Key Formula or Approach
Line 1: parallel to X-axis through \((1,4)\), so \(y-4=0\). Line 2: slope \(\tan45^{\circ}=1\) through \((1,4)\), so \(y-4=x-1\), i.e. \(x-y+3=0\).
Step 3: Detailed Explanation
\[ (y-4)(x-y+3)=0 \]
\[ xy-y^2+3y-4x+4y-12=0 \]
\[ xy-y^2-4x+7y-12=0 \]
Final Answer:
The joint equation is \(xy-y^2-4x+7y-12=0\), option (B). \[ \boxed{xy-y^2-4x+7y-12=0\ \text{(B)}} \]
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