Physics and Mathematics
Image of a Point in Different Cases
1. Concept Overview
The image of a point with respect to a line is the point obtained by reflecting the given point across the line, such that:
- The given line is the perpendicular bisector of the segment joining the point and its image
- The point and its image are at equal perpendicular distances from the line
- The line acts like a plane mirror
2. Conceptual Explanation (Geometric Meaning)
Let:
- A point be [(x_1, y_1)]
- A line (mirror) be [ax + by + c = 0]
- The image of the point be [(x_2, y_2)]
Then:
- The segment joining [(x_1, y_1)] and [(x_2, y_2)] is perpendicular to the line
- The midpoint of the segment lies on the line
This gives two powerful mathematical conditions.
3. Mathematical Derivation (Most Important)
Step 1: Midpoint Lies on the Line
Midpoint of [(x_1, y_1)] and [(x_2, y_2)] is:
[\left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)]
Since midpoint lies on the line:
[a \dfrac{x_1 + x_2}{2} + b \dfrac{y_1 + y_2}{2} + c ][= 0]
Step 2: Line Joining the Points is Perpendicular to Mirror
Slope of line [ax + by + c = 0] is:
[\left(m = -\dfrac{a}{b}\right)]
Slope of line joining point and image:
[\left(\dfrac{y_2 – y_1}{x_2 – x_1}\right)]
Since they are perpendicular:
[\left(\dfrac{y_2 – y_1}{x_2 – x_1} \cdot \left( -\dfrac{a}{b} \right) = -1\right)]
4. Direct Formula (Exam-Oriented)
The coordinates of the image of [(x_1, y_1)] in the line [ax + by + c = 0] are:
[x = x_1 – \dfrac{2a(ax_1 + by_1 + c)}{a^2 + b^2},] [ y = y_1 – \dfrac{2b(ax_1 + by_1 + c)}{a^2 + b^2}]
5. Key Features and Observations
- Image lies on the normal to the line
- Distance of point from line = distance of image from line
- Formula works for any orientation of line
- If point lies on the line, image coincides with the point
- Reflection preserves distance but reverses direction
6. Conceptual Questions with Detailed Solutions
1. Why must the mirror line be the perpendicular bisector?
Reflection preserves distance.
Hence the mirror must divide the segment joining the point and its image into two equal halves at right angles.
2. Why does the image always lie on the normal to the line?
Shortest distance from a point to a line is along the normal.
Reflection reverses this shortest path.
3. What happens if the point lies on the mirror line?
Then [ax_1 + by_1 + c = 0], so the image coordinates remain unchanged.
4. Does reflection change slope or distance?
Slope may change, but distance from the mirror remains unchanged.
5. Why is the formula symmetric in a and b?
Because reflection depends on the normal direction, not on x or y independently.
7. FAQ / Common Misconceptions (Deep Exam Points)
1. Students confuse image with foot of perpendicular.
Foot lies on the mirror; image lies symmetrically across it.
2. Forgetting factor of 2 in formula.
The image is twice the perpendicular distance away from the foot.
3. Using slope method for vertical lines.
Always prefer the direct formula for general cases.
4. Mixing distance formula and image formula.
Distance finds magnitude; image finds coordinates.
5. Assuming reflection preserves coordinates sign-wise.
Only distance is preserved, not sign or direction.
8. Practice Questions with Step-by-Step Solutions
Question 1: Find the image of the point [(2, 3)] in the line [x = 0].
Step-by-Step Solution:
1. Equation of line:
[x = 0] ⇒ [1·x + 0·y + 0 = 0]
2. Identify coefficients:
[a = 1,][ b = 0,][ c = 0]
3. Use image formula:
[x’ ][= x_1 – \dfrac{2a(ax_1 + by_1 + c)}{a^2 + b^2}]
4. Substitute values:
[x’ = 2 − 2·(2)/1 = −2]
5. For y-coordinate:
[y’ = 3 − 0 = 3]
Answer: Image is [(-2, 3)].
Question 2: Find the image of the point [(4, −1)] in the line [y = 0].
Step-by-Step Solution:
1. Line equation:
[y = 0] ⇒ [0·x + 1·y + 0 = 0]
2. Coefficients:
[a = 0,][ b = 1,][ c = 0]
3. Apply formula:
[y’ = −(−1) = 1]
4. x-coordinate remains same:
[x’ = 4]
Answer: Image is [(4, 1)].
Question 3: Find the image of [(1, 2)] in the line [x − y = 0].
Step-by-Step Solution:
1. Line coefficients:
[a = 1,][ b = −1,][ c = 0]
2. Compute expression:
[ax_1 + by_1 + c ][= 1 − 2 = −1]
3. Denominator:
[a^2 + b^2 = 2]
4. Apply formula:
[x’ = 1 − 2·1·(−1)/2 = 2]
[y’ = 2 − 2·(−1)·(−1)/2 = 1]
Answer: Image is [(2, 1)].
Question 4: Find the image of [(3, 1)] in the line [2x + y − 5 = 0].
Step-by-Step Solution:
1. Coefficients:
[a = 2, b = 1, c = −5]
2. Compute:
[ax_1 + by_1 + c ][= 6 + 1 − 5 = 2]
3. Denominator:
[a^2 + b^2 = 5]
4. Apply formula:
[x’ = 3 − 2·2·2/5 = 11/5]
[y’ = 1 − 2·1·2/5 = 1/5]
Answer: Image is [(11/5, 1/5)].
Question 5: Find the image of [(−2, 4)] in the line [x + 2y − 3 = 0].
Step-by-Step Solution:
1. Coefficients:
[a = 1,][ b = 2,][ c = −3]
2. Compute:
[ax_1 + by_1 + c ][= −2 + 8 − 3 = 3]
3. Denominator:
[a^2 + b^2 = 5]
4. Image coordinates:
[x’ = −2 − 2·1·3/5 ][= −16/5]
[y’ = 4 − 2·2·3/5 ][= 8/5]
Answer: Image is [(−16/5, 8/5)].
Question 6: Find the image of [(1, −3)] in the line [3x − 4y + 2 = 0].
Step-by-Step Solution:
1. Coefficients:
[a = 3,][ b = −4,][ c = 2]
2. Compute:
[ax_1 + by_1 + c ][= 3 + 12 + 2 ][= 17]
3. Denominator:
[a^2 + b^2 = 25]
4. Image:
[x’ = 1 − 2·3·17/25 ][= −77/25]
[y’ = −3 − 2·(−4)·17/25 ][= 61/25]
Answer: Image is [(−77/25, 61/25)].
Question 7: Find the image of [(0, 0)] in the line [4x + 3y − 12 = 0].
Step-by-Step Solution:
1. Compute:
[ax_1 + by_1 + c ][= −12]
2. Denominator:
[a^2 + b^2 = 25]
3. Image:
[x’ = 0 − 2·4·(−12)/25 ][= 96/25]
[y’ = 0 − 2·3·(−12)/25 ][= 72/25]
Answer: Image is [(96/25, 72/25)].
Question 8: Find the image of [(2, −1)] in the line [y = x].
Step-by-Step Solution:
1. Line equation:
[x − y = 0]
2. Reflection in [y = x] swaps coordinates.
Answer: Image is [(-1, 2)].
Question 9: Find the image of [(5, 2)] in the line [2x − y + 1 = 0].
Step-by-Step Solution:
1. Compute:
[ax_1 + by_1 + c = 10 − 2 + 1 = 9]
2. Denominator:
[a^2 + b^2 = 5]
3. Image:
[x’ = 5 − 36/5 ][= −11/5]
[y’ = 2 + 18/5 ][= 28/5]
Answer: Image is [(−11/5, 28/5)].
Question 10: Find the image of [(a, b)] in the line [x = 0].
Step-by-Step Solution:
1. Reflection in y-axis reverses x-coordinate.
Answer: Image is [(-a, b)].
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