Coordinate Geometry — Practice

Question 1

In which quadrant does the point (−5, −2) lie?

Accepted Answer

Step 1 — Read off the signs of the coordinates.
  x = −5 → negative.
  y = −2 → negative.

Step 2 — Match to the quadrant table.
  Both negative → Quadrant III (bottom-left).

✦ Answer: Quadrant III.

Why? Quadrant III is defined as the region where both x and y are negative. A quick mental check: bottom-left of the plane.

Question 2

In which quadrant does the point (3, −7) lie?

Accepted Answer

Step 1 — Signs.
  x = +3 (positive), y = −7 (negative).

Step 2 — Match the table.
  Positive x, negative y → Quadrant IV (bottom-right).

✦ Answer: Quadrant IV.

Mental picture: from the origin, move 3 units right, then 7 units down — you land in the bottom-right region.

Question 3

In which quadrant does the point (−8, 4) lie?

Accepted Answer

Step 1 — Signs.
  x = −8 (negative), y = +4 (positive).

Step 2 — Match the table.
  Negative x, positive y → Quadrant II (top-left).

✦ Answer: Quadrant II.

Quadrant II is anti-clockwise from Quadrant I; it covers the top-left of the plane.

Question 4

Where does the point (0, −9) lie?

Accepted Answer

Step 1 — Check whether either coordinate is zero.
  x = 0, y = −9. The x-coordinate is zero.

Step 2 — Recall the rule.
  Any point with x = 0 lies on the y-axis.
  Any point with y = 0 lies on the x-axis.
  Points on the axes do NOT belong to any quadrant.

Step 3 — Locate it.
  y = −9 means 9 units below the origin.

✦ Answer: On the y-axis, 9 units below the origin.

Common mistake: students often assign this to 'Quadrant III' because y is negative. It is on the boundary, not in any quadrant.

Question 5

Where does the point (4, 0) lie?

Accepted Answer

Step 1 — Check coordinates.
  x = 4, y = 0. The y-coordinate is zero.

Step 2 — Apply the rule.
  y = 0 means the point lies on the x-axis.

Step 3 — Position.
  x = +4 means 4 units to the right of the origin.

✦ Answer: On the x-axis, 4 units to the right of the origin. Not in any quadrant.

Question 6

Reflect (3, −4) across the x-axis.

Accepted Answer

Step 1 — Recall the reflection rule for the x-axis.
  Reflection across the x-axis: (x, y) → (x, −y).
  The x stays the same; y flips sign.

Step 2 — Apply.
  (3, −4) → (3, −(−4)) → (3, 4).

✦ Answer: (3, 4) — Quadrant I.

Verification: the original point and its image are equidistant from the x-axis (4 units), one below and one above — exactly what a mirror across the x-axis produces.

Question 7

Reflect (−2, 5) across the y-axis.

Accepted Answer

Step 1 — Recall the y-axis reflection rule.
  (x, y) → (−x, y). The y stays the same; x flips sign.

Step 2 — Apply.
  (−2, 5) → (−(−2), 5) → (2, 5).

✦ Answer: (2, 5) — Quadrant I.

Mnemonic: 'reflect across Y → Y stays.' The mirror's the y-axis, so the y-coordinate is unchanged.

Question 8

Reflect (4, −7) through the origin.

Accepted Answer

Step 1 — Recall the origin-reflection rule.
  (x, y) → (−x, −y). Both signs flip.
  (Reflection through the origin = 180° rotation about O.)

Step 2 — Apply.
  (4, −7) → (−4, −(−7)) → (−4, 7).

✦ Answer: (−4, 7) — Quadrant II.

Geometric check: original is in Quadrant IV (+,−); image is in Quadrant II (−,+). They sit on opposite sides of O, exactly as a 180° rotation produces.

Question 9

What is the ordinate of the point (6, −3)?

Accepted Answer

Step 1 — Recall the terminology.
  In the ordered pair (x, y), x is the abscissa and y is the ordinate.

Step 2 — Identify y in (6, −3).
  y = −3.

✦ Answer: Ordinate = −3.

Mnemonic — 'O for ordinate, O for over (up-down)': the ordinate is the vertical coordinate.

Question 10

What is the abscissa of the point (−5, 8)?

Accepted Answer

Step 1 — Terminology.
  Abscissa = x-coordinate (horizontal distance from y-axis, signed).

Step 2 — Identify x in (−5, 8).
  x = −5.

✦ Answer: Abscissa = −5.

Mnemonic — 'A for abscissa, A for along (left-right)': the abscissa is the horizontal coordinate.

Question 11

Plot the points A(2, 3), B(−2, 3), C(−2, −3), D(2, −3). Name the figure ABCD.

Accepted Answer

Step 1 — Identify which pairs share a coordinate.
  A & B both have y = 3 → joined by a HORIZONTAL segment.
  C & D both have y = −3 → also horizontal.
  A & D both have x = 2 → joined by a VERTICAL segment.
  B & C both have x = −2 → also vertical.

Step 2 — Compute side lengths.
  AB = |2 − (−2)| = 4 units.
  CD = |2 − (−2)| = 4 units.
  AD = |3 − (−3)| = 6 units.
  BC = |3 − (−3)| = 6 units.

Step 3 — Check angles.
  All sides meet at right angles (horizontal meets vertical), so all four angles are 90°.

Step 4 — Classify.
  Two pairs of equal, parallel sides + four right angles → rectangle.
  Length 6 ≠ breadth 4, so it is NOT a square.

✦ Answer: ABCD is a rectangle with length 6 and breadth 4. Area = 24 sq units, perimeter = 20 units.

Question 12

A point P(a, b) lies on the x-axis. What is b? Where is P if a = −7?

Accepted Answer

Step 1 — Apply the rule for points on the x-axis.
  Any point on the x-axis has y-coordinate 0.
  Therefore b = 0.

Step 2 — Substitute a = −7.
  P = (−7, 0).

Step 3 — Locate.
  Negative x → left of the y-axis.
  P is 7 units to the LEFT of the origin, on the x-axis.

✦ Answer: b = 0; P = (−7, 0), 7 units left of the origin.

Question 13

Find the distance between A(−3, 5) and B(7, 5).

Accepted Answer

Step 1 — Observe that both points share the same y-coordinate.
  y = 5 for both → AB is a HORIZONTAL segment lying on the line y = 5.

Step 2 — Use the horizontal-distance formula.
  d = |x₂ − x₁| (absolute value because distance is non-negative).

Step 3 — Substitute.
  d = |7 − (−3)| = |7 + 3| = |10| = 10.

✦ Answer: AB = 10 units.

Why absolute value? Without |…|, we'd get +10 in one direction and −10 in the other. Distance has no direction.

Question 14

Find the distance between P(4, −2) and Q(4, 6).

Accepted Answer

Step 1 — Observe the shared coordinate.
  Both points have x = 4 → PQ is a VERTICAL segment on the line x = 4.

Step 2 — Vertical distance formula.
  d = |y₂ − y₁|.

Step 3 — Substitute.
  d = |6 − (−2)| = |6 + 2| = 8.

✦ Answer: PQ = 8 units.

Note: had we written 'd = 6 − (−2) = 8' without the absolute-value bars, an examiner can deduct half a mark. Always wrap with |…|.

Question 15

Plot P(3, 0), Q(0, 4), R(−3, 0), S(0, −4). What figure do they form? Find its area.

Accepted Answer

Step 1 — Recognise where each point lies.
  P(3, 0) and R(−3, 0): on the x-axis, 3 units on either side of the origin.
  Q(0, 4) and S(0, −4): on the y-axis, 4 units on either side of the origin.

Step 2 — Join them in order PQRS.
  This produces a quadrilateral whose vertices are on the axes — a 'kite'-like shape sitting symmetrically about both axes.

Step 3 — Use the diagonal lengths.
  Diagonal PR (along x-axis): length = 3 − (−3) = 6.
  Diagonal QS (along y-axis): length = 4 − (−4) = 8.
  The diagonals are PERPENDICULAR (one is x-axis, the other is y-axis) and BISECT each other (both pass through the origin which is their midpoint).

Step 4 — Classify.
  Perpendicular diagonals that bisect each other ⇒ rhombus.
  Since the diagonals are UNEQUAL (6 ≠ 8), it is not a square.

Step 5 — Compute area.
  Area of a rhombus = (1/2) × d₁ × d₂ = (1/2) × 6 × 8 = 24 sq units.

✦ Answer: PQRS is a rhombus of area 24 sq units.

Question 16

Reflect (2, −5) first across the x-axis, then across the y-axis. What are the final coordinates?

Accepted Answer

Step 1 — First reflection (across the x-axis).
  Rule: (x, y) → (x, −y).
  (2, −5) → (2, −(−5)) → (2, 5).

Step 2 — Second reflection (across the y-axis).
  Rule: (x, y) → (−x, y).
  (2, 5) → (−2, 5).

✦ Answer: Final point (−2, 5) — Quadrant II.

Deeper insight: composing reflections across the x-axis and y-axis is equivalent to reflecting through the origin in one step — same as (x, y) → (−x, −y). Check: (2, −5) → (−2, 5). ✓

Question 17

Without plotting, predict the quadrant of (a, b) given a > 0 and b < 0.

Accepted Answer

Step 1 — Translate the conditions.
  a > 0 → x-coordinate is positive.
  b < 0 → y-coordinate is negative.

Step 2 — Match to the quadrant table.
  (+, −) → Quadrant IV (bottom-right).

✦ Answer: Quadrant IV.

No plotting needed — the signs alone determine the quadrant.

Question 18

Three vertices of a rectangle are (2, 4), (2, −2), (8, −2). Find the fourth vertex and the area.

Accepted Answer

Step 1 — Identify which sides of the rectangle are given.
  Two of the given vertices share x = 2 → the LEFT vertical side runs from (2, −2) to (2, 4).
  Two of the given vertices share y = −2 → the BOTTOM horizontal side runs from (2, −2) to (8, −2).

Step 2 — Use the rectangle property: opposite sides are equal and parallel.
  The fourth vertex must share x with (8, −2) (so the RIGHT side is vertical, parallel to the left).
  The fourth vertex must share y with (2, 4) (so the TOP side is horizontal, parallel to the bottom).
  Therefore the fourth vertex = (8, 4).

Step 3 — Compute side lengths.
  Vertical side: |4 − (−2)| = 6.
  Horizontal side: |8 − 2| = 6.

Step 4 — Compute the area.
  Area = length × breadth = 6 × 6 = 36 sq units.

✦ Answer: Fourth vertex = (8, 4). Area = 36 sq units.

Bonus insight: since length = breadth = 6, this 'rectangle' is actually a SQUARE. A square is a special rectangle. Examiners reward this observation.

Question 19

How many points (x, y) with integer coordinates lie strictly inside the rectangle with vertices (0, 0), (5, 0), (5, 3) and (0, 3)?

Accepted Answer

Step 1 — Interpret 'strictly inside'.
  Strictly inside means excluding the boundary — points ON the rectangle's sides do NOT count.

Step 2 — Range of valid x-values.
  The rectangle spans x = 0 to x = 5. 'Strictly inside' means 0 < x < 5.
  Integer x's satisfying this: x ∈ {1, 2, 3, 4} → 4 values.

Step 3 — Range of valid y-values.
  The rectangle spans y = 0 to y = 3. 'Strictly inside' means 0 < y < 3.
  Integer y's satisfying this: y ∈ {1, 2} → 2 values.

Step 4 — Combine using the multiplication principle.
  Each valid x can pair with each valid y, independently.
  Total = 4 × 2 = 8.

✦ Answer: 8 integer-coordinate points strictly inside the rectangle.

List them (verification): (1,1), (1,2), (2,1), (2,2), (3,1), (3,2), (4,1), (4,2) — exactly 8 points. ✓

This problem previews Pick's Theorem (Class 11 Olympiad: area = interior lattice + boundary/2 − 1).

Question 20

A bug walks anti-clockwise along the perimeter of the rectangle with vertices (0, 0), (4, 0), (4, 3), (0, 3) at 1 unit/sec, starting at (0, 0). Where is the bug at t = 11 sec?

Accepted Answer

Step 1 — Compute the perimeter.
  P = 2(length + breadth) = 2(4 + 3) = 14 units.
  The bug walks at 1 unit/sec → completes a full loop every 14 seconds.

Step 2 — Distance walked in 11 seconds.
  d = speed × time = 1 × 11 = 11 units.
  Since 11 < 14, the bug has NOT completed a full loop yet.

Step 3 — Trace the path anti-clockwise from (0, 0).
  (0,0) → (4,0): 4 units along the bottom.   Cumulative time = 4 s.   At t = 4, bug is at (4, 0).
  (4,0) → (4,3): 3 units up the right side.   Cumulative time = 4 + 3 = 7 s.   At t = 7, bug is at (4, 3).
  (4,3) → (0,3): 4 units along the top (leftward).   Cumulative time = 7 + 4 = 11 s.   At t = 11, bug is at (0, 3).

Step 4 — Conclusion.
  At t = 11, the bug arrives exactly at the corner (0, 3).

✦ Answer: The bug is at (0, 3) — the top-left corner of the rectangle.

Follow-up: at t = 14, the bug returns to (0, 0). At t = 28, two full loops. This is a periodic motion problem.

Question 21

A drone is at (2, 5) km from a school. It reflects its position across the x-axis to drop a payload, then reflects its new position across the y-axis. Find its final location.

Accepted Answer

Step 1 — Treat the school as the origin O(0, 0).
  Drone starts at (2, 5): 2 km EAST and 5 km NORTH of the school.

Step 2 — First reflection (across the x-axis = the East-West axis).
  Rule: (x, y) → (x, −y).
  (2, 5) → (2, −5). The drone is now 2 km EAST and 5 km SOUTH of the school.

Step 3 — Second reflection (across the y-axis = the North-South axis).
  Rule: (x, y) → (−x, y).
  (2, −5) → (−2, −5). The drone is now 2 km WEST and 5 km SOUTH.

Step 4 — Interpret in real-world terms.
  Final coordinates (−2, −5) — Quadrant III.

✦ Answer: The drone ends up 2 km WEST and 5 km SOUTH of the school, in Quadrant III.

Real-world check: this is exactly how some delivery drones plan return paths — using reflection symmetry to simplify route planning.

Practice — Coordinate Geometry

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