Question:
If the image of \(P(2,3)\) in a line \(L\) is \(Q(4,5)\), find the image of \(R(0,0)\) in the same line \(L\).
If the image of \(P(2,3)\) in a line \(L\) is \(Q(4,5)\), find the image of \(R(0,0)\) in the same line \(L\).
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The reflection line of a point and its image is always their perpendicular bisector.
Updated On: Jun 9, 2026
- \((4,5)\)
- \((3,4)\)
- \((2,2)\)
- \((7,7)\)
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The Correct Option is D
Solution and Explanation
Concept:
The mirror line is the perpendicular bisector of the segment joining a point and its image.
Step 1: Find the equation of the mirror line. Midpoint of \(PQ\): \[ M=\left(\frac{2+4}{2},\frac{3+5}{2}\right) =(3,4) \] Slope of \(PQ\): \[ m=\frac{5-3}{4-2}=1 \] Hence slope of mirror line \[ =-1 \] Equation: \[ y-4=-(x-3) \] \[ x+y-7=0 \]
Step 2: Reflect the origin about \(x+y-7=0\). Using reflection formula, \[ x' = 0-\frac{2(1)(-7)}{1^2+1^2} = 7 \] \[ y' = 0-\frac{2(1)(-7)}{1^2+1^2} = 7 \] Therefore, \[ \boxed{(7,7)} \]
Step 1: Find the equation of the mirror line. Midpoint of \(PQ\): \[ M=\left(\frac{2+4}{2},\frac{3+5}{2}\right) =(3,4) \] Slope of \(PQ\): \[ m=\frac{5-3}{4-2}=1 \] Hence slope of mirror line \[ =-1 \] Equation: \[ y-4=-(x-3) \] \[ x+y-7=0 \]
Step 2: Reflect the origin about \(x+y-7=0\). Using reflection formula, \[ x' = 0-\frac{2(1)(-7)}{1^2+1^2} = 7 \] \[ y' = 0-\frac{2(1)(-7)}{1^2+1^2} = 7 \] Therefore, \[ \boxed{(7,7)} \]
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