Reflecting a point is just flipping a sign, and the distance from a point to a line has one tidy formula. Both come up in locus questions, where you describe all points obeying a rule.
Flip a sign; plug into the formula
• Reflection in the x-axis:(x,y)→(x,−y). In the y-axis:(x,y)→(−x,y). In the origin:(x,y)→(−x,−y). In the line y=x: swap them, (x,y)→(y,x).
• Distance from a point(x0,y0) to the line ax+by+c=0 is a2+b2∣ax0+by0+c∣.
• Locus = the set of all points satisfying a condition; turn the words into an equation in x and y and simplify.
📐Coordinate Plane ExplorerLive calculations as you move points
P (x, y)
x
,
y
Distance line: ax + by + c = 0
a
b
c
Mirror in:
Reflection in the x-axisRule: (x, y) → (x, −y)P(1, 2) → P′(1, -2)
Distance from P to 3x + 4y − 10 = 0|ax₀+by₀+c| / √(a²+b²) = |3·1 + 4·2 + (-10)| / √(9+16) = |1| / √250.2 units
Reflection is a sign flip / swap — no calculation needed. The mirror line (dashed pink) is the perpendicular bisector of PP′.
The distance from (0, 0) to the line 3x + 4y − 10 = 0 is 9+16∣−10∣= .
📝Practice Questions
Q1The reflection of (3, 4) in the x-axis is:
Q2The reflection of (3, 4) in the y-axis is:
Q3The reflection of (5, 2) in the origin is:
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Real exam questions — Reflection & Distance to a Line
2 question types · 8 solved examples from real SSC papers
Two quick-scoring topics. Reflection is pure rule-recall — memorise the five mirror transformations and most questions are answered in seconds. Distance-to-a-line runs off one formula, a2+b2∣ax1+by1+c∣, while locus questions just ask you to recognise a definition (perpendicular bisector, ellipse, hyperbola).
How to solve this type
Lock these mirror rules in: in the x-axis (x,y)→(x,−y); in the y-axis (x,y)→(−x,y); in the origin (x,y)→(−x,−y); in the line y=x(x,y)→(y,x); in y=−x(x,y)→(−y,−x). For a parallel mirror use the midpoint idea: reflection in x=k is (2k−x,y) and reflection in y=k is (x,2k−y) — the coordinate matching the mirror stays fixed, the other one jumps to the equal distance on the far side.
Find the reflection of the point (4,2) in the line x=3.
A(2,2)B(6,2)C(−2,2)D(4,4)
Find the reflection of the point (2,3) in the line x=y.
A(−3,−2)B(2,3)C(−2,−3)D(3,2)
Find the reflection of the point (3,4) in the origin.
A(4,3)B(3,4)C(−4,−3)D(−3,−4)
Find the reflection of the point (5,−3) in the line y=3.
