Exercise 7.1 — Distance Formula
Distance between two points in the coordinate plane.
Loading PDF…
What is Co-ordinate Geometry?
Co-ordinate Geometry (also called Analytical Geometry) is the branch of mathematics that connects algebra and geometry. It allows us to describe the exact position of any point in a plane using a pair of numbers called coordinates — written as (x, y). The horizontal line is called the x-axis and the vertical line is the y-axis; together they form the Cartesian plane, named after the French mathematician René Descartes.
In Class 10 (CBSE, Telangana, and Andhra Pradesh syllabus), Chapter 7 covers three major tools: the Distance Formula, the Section Formula, and the Area of a Triangle using coordinates. Exercise 7.1 focuses entirely on the Distance Formula — the most foundational skill of this chapter.
💡 Why learn this? The distance formula is used in real life for GPS navigation, map distances, engineering, architecture, and even in data science. Mastering it builds strong foundations for board exam problems worth 4–6 marks.
The Distance Formula — How It Is Derived
The distance formula is not just a rule to memorise — it is derived from the Pythagoras Theorem. Understanding the derivation helps you remember and apply it correctly even in unfamiliar problems.
Right triangle △ABC formed by two points A and B
∠C = 90° — Pythagoras theorem applies
Given two points A(x₁, y₁) and B(x₂, y₂) in the coordinate plane:
- Drop perpendiculars from A and B to the x-axis at points P and Q respectively.
- Let C be the foot of perpendicular from A onto BQ.
- The horizontal leg AC = PQ = x₂ − x₁
- The vertical leg BC = BQ − CQ = y₂ − y₁
- △ABC is a right-angled triangle with ∠C = 90°.
- Applying Pythagoras Theorem: AB² = AC² + BC²
📐 Distance Formula — The Core of Exercise 7.1
AB = √[ (x₂ − x₁)² + (y₂ − y₁)² ]
Special case: Distance from Origin O(0,0) to point P(x, y) = √(x² + y²)
📌 Important: The formula works regardless of which point is labelled (x₁, y₁) or (x₂, y₂) — since both differences are squared, the negative sign disappears. So you can swap the two points and still get the same answer.
Question 1 — Distance Between Pairs of Points
Find the distance between the following pairs of points.
Q1 (i)
Distance between (2, 3) and (4, 1)
Here x₁ = 2, y₁ = 3, x₂ = 4, y₂ = 1. Substitute into the distance formula.
d = √[ (x₂−x₁)² + (y₂−y₁)² ]
= √[ (4−2)² + (1−3)² ]
= √[ (2)² + (−2)² ]
= √[ 4 + 4 ]
= √8 = √(4 × 2) = 2√2 units
Answer: 2√2 units ≈ 2.83 units
Q1 (ii)
Distance between (−5, 7) and (−1, 3)
Here x₁ = −5, y₁ = 7, x₂ = −1, y₂ = 3. Be careful with the negative signs!
d = √[ (−1−(−5))² + (3−7)² ]
= √[ (−1+5)² + (−4)² ]
= √[ (4)² + (−4)² ]
= √[ 16 + 16 ]
= √32 = √(16 × 2) = 4√2 units
Answer: 4√2 units ≈ 5.66 units
Q1 (iii)
Distance between (−2, −3) and (3, 2)
Both points are in different quadrants. Subtracting a negative value adds to the result.
d = √[ (3−(−2))² + (2−(−3))² ]
= √[ (3+2)² + (2+3)² ]
= √[ (5)² + (5)² ]
= √[ 25 + 25 ]
= √50 = √(25 × 2) = 5√2 units
Answer: 5√2 units ≈ 7.07 units
Q1 (iv)
Distance between (a, b) and (−a, −b)
This is an algebraic question — the answer is in terms of a and b. These two points are symmetric about the origin.
d = √[ (−a−a)² + (−b−b)² ]
= √[ (−2a)² + (−2b)² ]
= √[ 4a² + 4b² ]
= √[ 4(a² + b²) ]
= √4 × √(a²+b²) = 2√(a²+b²) units
Answer: 2√(a² + b²) units
💡 Key pattern in Q1(iv): The points (a, b) and (−a, −b) are mirror images of each other through the origin. The distance between any point and its "origin-reflection" is always 2√(a²+b²) — twice the distance from the point to the origin.
Question 2 — Distance from the Origin
Q2
Distance between (0, 0) and (36, 15)
One point is the origin, so use the simplified formula: d = √(x² + y²)
d = √(x² + y²)
= √(36² + 15²)
= √(1296 + 225)
= √1521
= 39 units (since 39² = 1521 — verify: 39×39 = 1521 ✓)
Answer: 39 units (exact whole number!)
📌 How to check √1521 = 39: Prime factorisation method — 1521 = 3 × 507 = 3 × 3 × 169 = 9 × 169 = 9 × 13² = (3×13)² = 39². So √1521 = 39. Always try to check if the number under the root is a perfect square first!
Advertisement
Question 3 — Testing for Collinearity
Three points are collinear (lying on the same straight line) if and only if the sum of the two shorter distances equals the longest distance. In other words, if A, B, C are collinear, then AB + BC = AC (or some arrangement of this). If this condition fails, the points are not collinear.
Q3
Verify whether (1, 5), (2, 3) and (−2, −1) are collinear
AB (A(1,5) to B(2,3))
= √[(2−1)²+(3−5)²]
= √[1+4] = √5 units
BC (B(2,3) to C(−2,−1))
= √[(−2−2)²+(−1−3)²]
= √[16+16] = √32 units
AC (A(1,5) to C(−2,−1))
= √[(−2−1)²+(−1−5)²] = √[9+36] = √45 = 3√5 units
Check: AB + BC = √5 + √32 ≈ 2.24 + 5.66 = 7.90
AC = 3√5 ≈ 6.71
AB + BC ≠ AC → Points are NOT collinear
⛔ Common mistake: Students sometimes check only one pair (AB + BC) without checking all three combinations. Make sure you identify the largest distance first and check whether the other two add up to it.
Identifying Triangle and Quadrilateral Types Using Distances
A powerful application of the distance formula is identifying the type of geometric figure formed by a set of points. The table below summarises the conditions for each shape — this is essential knowledge for Questions 4 through 9.
| Shape | Condition on Sides | Additional Check |
|---|---|---|
| Equilateral Triangle | All 3 sides equal (AB = BC = CA) | — |
| Isosceles Triangle | Exactly 2 sides equal | — |
| Scalene Triangle | All 3 sides different | — |
| Right-angled Triangle | AB² + BC² = AC² (Pythagoras) | Longest side is hypotenuse |
| Parallelogram | Opposite sides equal: AB = CD, BC = AD | Diagonals AC ≠ BD (unequal) |
| Rectangle | Opposite sides equal: AB = CD, BC = AD | Diagonals AC = BD (equal) |
| Rhombus | All 4 sides equal: AB = BC = CD = AD | Diagonals AC ≠ BD (unequal) |
| Square | All 4 sides equal: AB = BC = CD = AD | Diagonals AC = BD (equal) |
💡 Board Exam Tip: When asked to "name the quadrilateral," always check both the sides AND the diagonals. Equal sides + equal diagonals = Square. Equal sides + unequal diagonals = Rhombus. Opposite sides equal + equal diagonals = Rectangle. Opposite sides equal + unequal diagonals = Parallelogram.
Questions 4, 5 & 6 — Triangle Type Problems
Q4
Check whether (5, −2), (6, 4) and (7, −2) are vertices of an isosceles triangle
AB
= √[(6−5)²+(4−(−2))²]
= √[1+36] = √37 units
BC
= √[(7−6)²+(−2−4)²]
= √[1+36] = √37 units
AC
= √[(7−5)²+(−2−(−2))²]
= √[4+0] = 2 units
Conclusion:
AB = BC = √37
AB ≠ AC
✅ Since exactly two sides are equal (AB = BC), the given points form an isosceles triangle.
Q5
Classroom Problem — Is ABCD a square? A(3,4), B(6,7), C(9,4), D(6,1)
Jarina claims ABCD is a square. Phani disagrees. Use the distance formula to settle it.
AB = √[(6−3)²+(7−4)²]
= √[9+9] = √18
BC = √[(9−6)²+(4−7)²]
= √[9+9] = √18
CD = √[(6−9)²+(1−4)²]
= √[9+9] = √18
DA = √[(3−6)²+(4−1)²]
= √[9+9] = √18
Diagonal AC = √[(9−3)²+(4−4)²] = √[36+0] = 6 units
Diagonal BD = √[(6−6)²+(1−7)²] = √[0+36] = 6 units
All 4 sides equal (√18) AND diagonals equal (6 = 6) → ABCD is a SQUARE ✓
✅ Jarina is correct! ABCD satisfies both conditions for a square: all sides equal and both diagonals equal.
Q6
Show that A(a, 0), B(−a, 0) and C(0, a√3) form an equilateral triangle
AB = √[(−a−a)²+(0−0)²]
= √[(−2a)²] = √(4a²) = 2a
BC = √[(0−(−a))²+(a√3−0)²]
= √[a²+3a²] = √(4a²) = 2a
CA = √[(a−0)²+(0−a√3)²]
= √[a²+3a²] = √(4a²) = 2a
AB = BC = CA = 2a
✅ Since all three sides are equal (each = 2a), the given points form an equilateral triangle.
Questions 7 & 8 — Parallelogram and Rhombus Problems
Q7
Prove that (−7,−3), (5,10), (15,8) and (3,−5) form a parallelogram
Label A(−7,−3), B(5,10), C(15,8), D(3,−5). For a parallelogram, opposite sides must be equal.
AB = √[(5+7)²+(10+3)²]
= √[144+169] = √313
CD = √[(3−15)²+(−5−8)²]
= √[144+169] = √313
BC = √[(15−5)²+(8−10)²]
= √[100+4] = √104
AD = √[(3+7)²+(−5+3)²]
= √[100+4] = √104
✅ AB = CD = √313 and BC = AD = √104. Opposite sides are equal, so ABCD is a parallelogram.
Q8
Show that (−4,−7), (−1,2), (8,5) and (5,−4) are vertices of a rhombus and find its area
AB = √[(3)²+(9)²] = √90
BC = √[(9)²+(3)²] = √90
CD = √[(−3)²+(−9)²] = √90
AD = √[(9)²+(3)²] = √90
Diagonal AC
= √[(12)²+(12)²] = √288 = 12√2
Diagonal BD
= √[(6)²+(−6)²] = √72 = 6√2
AC ≠ BD → Rhombus (not square)
Area of Rhombus = ½ × d₁ × d₂ = ½ × 12√2 × 6√2
= ½ × 72 × 2 = ½ × 144 = 72 sq. units
✅ All 4 sides equal (√90) but diagonals unequal → Rhombus. Area = 72 sq. units.
Question 9 — Name the Type of Quadrilateral
Q9 (i)
(−1,−2), (1,0), (−1,2), (−3,0) — What quadrilateral?
AB = √[(1+1)²+(0+2)²] = √[4+4] = √8
BC = √[(−1−1)²+(2−0)²] = √[4+4] = √8
CD = √[(−3+1)²+(0−2)²] = √[4+4] = √8
AD = √[(−3+1)²+(0+2)²] = √[4+4] = √8 → all sides equal
Diagonal AC = √[(−1+1)²+(2+2)²] = √16 = 4
Diagonal BD = √[(−3−1)²+(0−0)²] = √16 = 4
All sides equal AND diagonals equal → SQUARE ✓
Answer: Square
Q9 (ii)
(−3,5), (3,1), (1,−3), (−5,1) — What quadrilateral?
AB = √[(6)²+(−4)²] = √52 | BC = √[(−2)²+(−4)²] = √20
CD = √[(−6)²+(4)²] = √52 | AD = √[(−2)²+(−4)²] = √20
Diagonal AC = √[16+64] = √80
Diagonal BD = √[64+0] = 8
Opposite sides equal (AB=CD, BC=AD), BUT AC ≠ BD → PARALLELOGRAM ✓
Answer: Parallelogram
Q9 (iii)
(4,5), (7,6), (4,3), (1,2) — What quadrilateral?
AB = √[(3)²+(1)²] = √10 | BC = √[(−3)²+(−3)²] = √18
CD = √[(−3)²+(1)²] = √10 | AD = √[(−3)²+(−3)²] = √18
Diagonal AC = √[0+4] = 2
Diagonal BD = √[36+16] = √52
Opposite sides equal (AB=CD=√10, BC=AD=√18), BUT AC ≠ BD → PARALLELOGRAM ✓
Answer: Parallelogram
Questions 10, 11 & 12 — Finding Unknown Coordinates
These questions reverse the usual process — instead of finding distance, you are given the distance (or a condition) and asked to find an unknown coordinate. The strategy is to set up an equation using the distance formula and solve for the unknown.
Q10
Find the point on the X-axis equidistant from (2,−5) and (−2,9)
Any point on the x-axis has y-coordinate = 0. Let the required point be P(x, 0).
PA = PB (equidistant condition)
√[(x−2)²+(0+5)²] = √[(x+2)²+(0−9)²]
Squaring both sides:
(x−2)² + 25 = (x+2)² + 81
x²−4x+4+25 = x²+4x+4+81
−4x + 29 = 4x + 85
−8x = 56 → x = −7
Answer: P(−7, 0)
Q11
If distance between (x, 7) and (1, 15) is 10, find x
√[(x−1)²+(7−15)²] = 10
√[(x−1)²+64] = 10
Squaring: (x−1)²+64 = 100
(x−1)² = 36
x²−2x+1−36 = 0 → x²−2x−35 = 0
Using quadratic formula: x = [2 ± √(4+140)] / 2 = [2 ± 12] / 2
x = 14/2 = 7 OR x = −10/2 = −5
Answer: x = 7 or x = −5
Q12
Distance between P(2,−3) and Q(10,y) is 10 units. Find y.
√[(10−2)²+(y+3)²] = 10
√[64+(y+3)²] = 10
Squaring: 64 + (y+3)² = 100
(y+3)² = 36
y²+6y+9−36 = 0 → y²+6y−27 = 0
y = [−6 ± √(36+108)] / 2 = [−6 ± 12] / 2
y = 6/2 = 3 OR y = −18/2 = −9
Answer: y = 3 or y = −9
Questions 13, 14 & 15 — Circle Radius, Triangle Validity, and Equidistance Relation
Q13
Radius of circle with centre (3,2) passing through (−5,6)
The radius is simply the distance from the centre to any point on the circle.
r = √[(−5−3)²+(6−2)²]
= √[(−8)²+(4)²]
= √[64+16] = √80
= √(16×5) = 4√5 units
Answer: Radius = 4√5 units ≈ 8.94 units
Q14
Can a triangle be drawn with vertices (1,5), (5,8) and (13,14)?
A triangle can be formed only if the sum of any two sides is strictly greater than the third side.
AB = √[(4)²+(3)²] = √25 = 5 units
BC = √[(8)²+(6)²] = √100 = 10 units
AC = √[(12)²+(9)²] = √225 = 15 units
Check: AB + BC = 5 + 10 = 15 = AC
Sum of two sides = third side (NOT greater than) → Points are collinear!
⛔ No triangle can be formed. Since AB + BC = AC, the three points are collinear — they lie on the same straight line and cannot form a triangle.
Q15
Find the relation between x and y if (x,y) is equidistant from (−2,8) and (−3,−5)
PA = PB → PA² = PB² (squaring both sides)
(x+2)² + (y−8)² = (x+3)² + (y+5)²
x²+4x+4 + y²−16y+64 = x²+6x+9 + y²+10y+25
4x − 16y + 68 = 6x + 10y + 34
4x − 6x − 16y − 10y = 34 − 68
−2x − 26y = −34
Divide by −2: x + 13y = 17
Answer: x + 13y = 17
📌 This result represents a straight line — every point on the line x + 13y = 17 is equidistant from (−2,8) and (−3,−5). This line is called the perpendicular bisector of the segment joining those two points.
Quick Reference — All 15 Answers at a Glance
| Q# | Problem | Answer | Shape / Result |
|---|---|---|---|
| 1(i) | (2,3) and (4,1) | 2√2 units | — |
| 1(ii) | (−5,7) and (−1,3) | 4√2 units | — |
| 1(iii) | (−2,−3) and (3,2) | 5√2 units | — |
| 1(iv) | (a,b) and (−a,−b) | 2√(a²+b²) | Algebraic result |
| 2 | (0,0) and (36,15) | 39 units | Exact integer! |
| 3 | (1,5),(2,3),(−2,−1) | Not collinear | AB+BC ≠ AC |
| 4 | (5,−2),(6,4),(7,−2) | Isosceles △ | AB = BC = √37 |
| 5 | A(3,4),B(6,7),C(9,4),D(6,1) | Square | Jarina is correct |
| 6 | (a,0),(−a,0),(0,a√3) | Equilateral △ | All sides = 2a |
| 7 | (−7,−3),(5,10),(15,8),(3,−5) | Parallelogram | AB=CD=√313 |
| 8 | (−4,−7),(−1,2),(8,5),(5,−4) | Rhombus, 72 sq.u | All sides = √90 |
| 9(i) | (−1,−2),(1,0),(−1,2),(−3,0) | Square | All sides = √8, diags = 4 |
| 9(ii) | (−3,5),(3,1),(1,−3),(−5,1) | Parallelogram | Opposite sides equal |
| 9(iii) | (4,5),(7,6),(4,3),(1,2) | Parallelogram | Opposite sides equal |
| 10 | X-axis, equidist. from (2,−5)&(−2,9) | (−7, 0) | — |
| 11 | (x,7) to (1,15) = 10 | x = 7 or −5 | Quadratic |
| 12 | P(2,−3) to Q(10,y) = 10 | y = 3 or −9 | Quadratic |
| 13 | Circle centre (3,2), through (−5,6) | r = 4√5 units | — |
| 14 | (1,5),(5,8),(13,14) | No triangle | Collinear points |
| 15 | (x,y) equidist. from (−2,8)&(−3,−5) | x + 13y = 17 | Perpendicular bisector |
Common Mistakes to Avoid in Exams
- Sign errors with negative coordinates: When x₁ or y₁ is negative, subtracting it adds to the value. For example, 3 − (−2) = 3 + 2 = 5, NOT 1. This is the most common source of wrong answers in this exercise.
- Forgetting to check diagonals for quadrilaterals: Always compute both diagonals. Equal sides + equal diagonals = Square. Equal sides + unequal diagonals = Rhombus. Never conclude a shape is a square without checking diagonals.
- Stopping at "opposite sides equal" and calling it a Rectangle: Opposite sides equal AND diagonals equal = Rectangle. Just opposite sides equal = Parallelogram.
- Not simplifying surds: √8 must be simplified to 2√2, and √50 to 5√2. Board examiners deduct marks for unsimplified surds.
- Squaring only one side in equidistance problems: In Q10–Q12, you must square both sides of the equation before expanding. Missing this step leads to errors in the algebra.
- Confusing collinear check: For collinearity, AB + BC = AC (the largest), not any random arrangement. Identify the largest side first!
⛔ High-risk exam trap: Question 14 tests a concept students often overlook — when AB + BC = AC (not greater than AC), the points are collinear and no triangle is formed. Always check this condition before claiming a triangle exists.
Board Exam Tips and What This Exercise Prepares You For
Exercise 7.1 builds the foundation you need for the rest of Chapter 7. Once you are comfortable with the Distance Formula, the Section Formula (Exercise 7.2) — which divides a line segment in a given ratio — becomes much more approachable. The ability to compute distances also connects directly to the Area of a Triangle using coordinates (Exercise 7.3).
Understanding how to classify triangles and quadrilaterals using coordinates also reinforces concepts from Chapter 6 (Triangles) and Class 9 Co-ordinate Geometry. In Telangana and Andhra Pradesh board exams, Exercise 7.1 problems regularly appear in the 4-mark section, particularly shape identification problems (Q4–Q9) and finding unknown coordinates (Q10–Q12).
📐 Telangana & AP Board Exam Tip: Questions 5 (classroom/square), 8 (rhombus + area), and 10 (equidistant point on X-axis) are the most frequently examined problems from Exercise 7.1. Practise these thoroughly. In CBSE board papers, Q11 and Q12 (finding unknown coordinate given a distance) are perennial favourites for 3–4 mark questions.
📌 Distance Formula
✅ Collinearity Test
🔺 Triangle Types
▱ Quadrilateral Classification
⭕ Circle Radius
📝 Equidistance Problems