Question:
The points on the line \(x+y-4=0\) which are at one unit distance from the line \(4x+3y-10=0\) are...
The points on the line \(x+y-4=0\) which are at one unit distance from the line \(4x+3y-10=0\) are...
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Write a general point of the first line and set its distance from the second line equal to one.
Updated On: Nov 12, 2026
- \((5,-1)\) and \((-2,6)\)
- \((3,1)\) and \((-7,11)\)
- \((9,-5)\) and \((-11,15)\)
- \((-4,8)\) and \((2,2)\)
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The Correct Option is B
Solution and Explanation
Step 1: Set Up
A point on \(x+y-4=0\) is \((x,\,4-x)\).
Step 2: Apply the Distance Formula
Distance from \(4x+3y-10=0\) is \(\dfrac{|4x+3(4-x)-10|}{\sqrt{16+9}}=\dfrac{|x+2|}{5}\).
Step 3: Solve
\(\dfrac{|x+2|}{5}=1\) gives \(x+2=\pm 5\), so \(x=3\) or \(x=-7\).
The points are \((3,1)\) and \((-7,11)\), option (B).
Final Answer:
The points are \((3,1)\) and \((-7,11)\). \[ \boxed{\text{(B) }(3,1)\text{ and }(-7,11)} \]
A point on \(x+y-4=0\) is \((x,\,4-x)\).
Step 2: Apply the Distance Formula
Distance from \(4x+3y-10=0\) is \(\dfrac{|4x+3(4-x)-10|}{\sqrt{16+9}}=\dfrac{|x+2|}{5}\).
Step 3: Solve
\(\dfrac{|x+2|}{5}=1\) gives \(x+2=\pm 5\), so \(x=3\) or \(x=-7\).
The points are \((3,1)\) and \((-7,11)\), option (B).
Final Answer:
The points are \((3,1)\) and \((-7,11)\). \[ \boxed{\text{(B) }(3,1)\text{ and }(-7,11)} \]
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