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Distance between two points · 2 marks

1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis?
2. Can these distances help you find the distance AD?

  1. 1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis?
  2. 2. Can these distances help you find the distance AD?
Answer: 1. Along the x-axis: 7 − 3 = 4 units (to the right); along the y-axis: 4 − 1 = 3 units (downwards). 2. Yes: they are the legs of a right-angled triangle with hypotenuse AD, so AD = √(42 + 32) = 5 units.

Step-by-step solution

Idea: Any move from one point to another can be split into a move parallel to the x-axis and a move parallel to the y-axis. These two moves are at right angles.

xy−1123456789101234567C (3, 1)435A (3, 4)D (7, 1)M (9, 6)

1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis?

  1. Along the x-axis the x-coordinate changes from 3 to 7: distance = 7 − 3 = 4 units (to the right).½ mark
  2. Along the y-axis the y-coordinate changes from 4 to 1: distance = 4 − 1 = 3 units (downwards).½ mark
4 units along the x-axis and 3 units along the y-axis.

2. Can these distances help you find the distance AD?

  1. Yes. Go from A straight down 3 units to C (3, 1), then 4 units across to D. AC is vertical and CD is horizontal, so the angle at C is a right angle, and AD is the hypotenuse of triangle ACD.½ mark
  2. By the Baudhāyana–Pythagoras theorem, AD = √(CD2 + AC2) = √(42 + 32) = √25 = 5 units.½ mark
Yes: AD = √(42 + 32) = 5 units.
1. 4 units along the x-axis and 3 units along the y-axis. 2. Yes: they are the legs of a right-angled triangle with hypotenuse AD, so AD = √(4² + 3²) = 5 units.

Check: Count on the grid: from A go 3 squares down and 4 squares right to reach D ✓. 3, 4, 5 is a Pythagorean triple: 9 + 16 = 25 ✓.

Answer to write in the exam

1.

Distance along the x-axis = 7 − 3 = 4 units

Distance along the y-axis = 4 − 1 = 3 units

2.

C (3, 1): AC = 3, CD = 4, ∠ACD = 90°

AD2 = CD2 + AC2 (Baudhāyana–Pythagoras theorem) = 16 + 9 = 25

∴ AD = 5 units

Common mistakes that cost marks

  • Writing the y-distance as 1 − 4 = −3 units. A distance is never negative; the shift is 3 units (downwards).
  • Thinking AD = 4 + 3 = 7. That is the length of the path A → C → D, not the straight line AD.
  • Subtracting the wrong coordinates, e.g. 7 − 4 = 3 (an x-coordinate minus a y-coordinate).

How this can come in the exam

MCQ (1 mark)

A point moves from (−2, 5) to (4, −3). The distances covered along the x-axis and the y-axis are

  1. 6 and 8
  2. 2 and 2
  3. 6 and 2
  4. 2 and 8
Show answer

(A) 6 and 8
4 − (−2) = 6 and 5 − (−3) = 8.

Assertion–Reason (1 mark)

Assertion (A): The distance between (−3, 5) and (4, −19) is 25 units.
Reason (R): The horizontal and vertical shifts between two points are the legs of a right-angled triangle whose hypotenuse is the distance between them.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Show answer

(A) Both A and R are true, and R is the correct explanation of A.
Shifts 7 and 24; by R the distance is √(49 + 576) = √625 = 25.

Try one yourself

In moving from P (1, 1) to Q (3, 5), what distances are covered along the two axes? Hence find PQ.

Show answer

Along the x-axis 2 units, along the y-axis 4 units. PQ = √(4 + 16) = √20 = 2√5 units.

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