April 10, 2026
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Chapter 7: Coordinate Geometry
Complete NCERT Solutions | CBSE Class 10 Mathematics (2025-26)
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Exercise 7.1 (Page 161)
12 marks
Find the distance between the following pairs of points :
(i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (-a, -b)
(i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (-a, -b)
Answer:
(i) Distance = √[(4-2)² + (1-3)²] = √[4 + 4] = √8 = 2√2 units
(ii) Distance = √[(-1+5)² + (3-7)²] = √[16 + 16] = √32 = 4√2 units
(iii) Distance = √[(-a-a)² + (-b-b)²] = √[(-2a)² + (-2b)²] = √[4a²+4b²] = 2√(a²+b²) units
(i) Distance = √[(4-2)² + (1-3)²] = √[4 + 4] = √8 = 2√2 units
(ii) Distance = √[(-1+5)² + (3-7)²] = √[16 + 16] = √32 = 4√2 units
(iii) Distance = √[(-a-a)² + (-b-b)²] = √[(-2a)² + (-2b)²] = √[4a²+4b²] = 2√(a²+b²) units
22 marks
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?
Answer:
Distance = √[(36-0)² + (15-0)²] = √[1296 + 225] = √1521 = 39 km
Yes, the distance between towns A and B is 39 km.
Distance = √[(36-0)² + (15-0)²] = √[1296 + 225] = √1521 = 39 km
Yes, the distance between towns A and B is 39 km.
33 marks
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Answer:
AB = √[(2-1)² + (3-5)²] = √[1+4] = √5
BC = √[(-2-2)² + (-11-3)²] = √[16+196] = √212 = 2√53
AC = √[(-2-1)² + (-11-5)²] = √[9+256] = √265
Since AB + BC ≠ AC, points are not collinear.
AB = √[(2-1)² + (3-5)²] = √[1+4] = √5
BC = √[(-2-2)² + (-11-3)²] = √[16+196] = √212 = 2√53
AC = √[(-2-1)² + (-11-5)²] = √[9+256] = √265
Since AB + BC ≠ AC, points are not collinear.
43 marks
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Answer:
AB = √[(6-5)² + (4+2)²] = √[1+36] = √37
BC = √[(7-6)² + (-2-4)²] = √[1+36] = √37
AC = √[(7-5)² + (-2+2)²] = √[4+0] = 2
Since AB = BC, triangle is isosceles.
AB = √[(6-5)² + (4+2)²] = √[1+36] = √37
BC = √[(7-6)² + (-2-4)²] = √[1+36] = √37
AC = √[(7-5)² + (-2+2)²] = √[4+0] = 2
Since AB = BC, triangle is isosceles.
53 marks
In a classroom, 4 friends are seated at points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think ABCD is a square?" Chameli disagrees. Using distance formula, find which of them is correct.
Answer:
From figure, coordinates: A(3,4), B(6,7), C(9,4), D(6,1).
AB = √[(6-3)²+(7-4)²] = √[9+9] = √18 = 3√2
BC = √[(9-6)²+(4-7)²] = √[9+9] = 3√2
CD = √[(6-9)²+(1-4)²] = √[9+9] = 3√2
DA = √[(3-6)²+(4-1)²] = √[9+9] = 3√2
AC = √[(9-3)²+(4-4)²] = √36 = 6
BD = √[(6-6)²+(1-7)²] = √36 = 6
All sides equal and diagonals equal → Champa is correct (square).
From figure, coordinates: A(3,4), B(6,7), C(9,4), D(6,1).
AB = √[(6-3)²+(7-4)²] = √[9+9] = √18 = 3√2
BC = √[(9-6)²+(4-7)²] = √[9+9] = 3√2
CD = √[(6-9)²+(1-4)²] = √[9+9] = 3√2
DA = √[(3-6)²+(4-1)²] = √[9+9] = 3√2
AC = √[(9-3)²+(4-4)²] = √36 = 6
BD = √[(6-6)²+(1-7)²] = √36 = 6
All sides equal and diagonals equal → Champa is correct (square).
63 marks
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Answer:
(i) Square (all sides = 2√2, diagonals = 4, perpendicular)
(ii) No special quadrilateral (not a parallelogram as diagonals don't bisect)
(iii) Parallelogram (midpoints of diagonals coincide)
(i) Square (all sides = 2√2, diagonals = 4, perpendicular)
(ii) No special quadrilateral (not a parallelogram as diagonals don't bisect)
(iii) Parallelogram (midpoints of diagonals coincide)
73 marks
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Answer:
Let point be (x, 0). Then (x-2)² + (0+5)² = (x+2)² + (0-9)²
x² - 4x + 4 + 25 = x² + 4x + 4 + 81
-4x + 29 = 4x + 85 → -8x = 56 → x = -7
Point: (-7, 0)
Let point be (x, 0). Then (x-2)² + (0+5)² = (x+2)² + (0-9)²
x² - 4x + 4 + 25 = x² + 4x + 4 + 81
-4x + 29 = 4x + 85 → -8x = 56 → x = -7
Point: (-7, 0)
83 marks
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Answer:
(10-2)² + (y+3)² = 100 → 64 + (y+3)² = 100 → (y+3)² = 36
y+3 = ±6 → y = 3 or y = -9
(10-2)² + (y+3)² = 100 → 64 + (y+3)² = 100 → (y+3)² = 36
y+3 = ±6 → y = 3 or y = -9
93 marks
If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Answer:
PQ² = (5-0)² + (-3-1)² = 25+16=41
RQ² = (x-0)² + (6-1)² = x²+25 = 41 → x² = 16 → x = ±4
For x=4: R(4,6), QR = √41, PR = √[(4-5)²+(6+3)²] = √[1+81]=√82
For x=-4: R(-4,6), PR = √[(-4-5)²+(6+3)²] = √[81+81]=9√2
PQ² = (5-0)² + (-3-1)² = 25+16=41
RQ² = (x-0)² + (6-1)² = x²+25 = 41 → x² = 16 → x = ±4
For x=4: R(4,6), QR = √41, PR = √[(4-5)²+(6+3)²] = √[1+81]=√82
For x=-4: R(-4,6), PR = √[(-4-5)²+(6+3)²] = √[81+81]=9√2
103 marks
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
Answer:
(x-3)² + (y-6)² = (x+3)² + (y-4)²
x²-6x+9 + y²-12y+36 = x²+6x+9 + y²-8y+16
-6x-12y+45 = 6x-8y+25 → -12x -4y +20 = 0 → 3x + y - 5 = 0
(x-3)² + (y-6)² = (x+3)² + (y-4)²
x²-6x+9 + y²-12y+36 = x²+6x+9 + y²-8y+16
-6x-12y+45 = 6x-8y+25 → -12x -4y +20 = 0 → 3x + y - 5 = 0
Exercise 7.2 (Page 167) – Section Formula
12 marks
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.
Answer:
x = [2×4 + 3×(-1)]/(2+3) = (8-3)/5 = 1
y = [2×(-3) + 3×7]/5 = (-6+21)/5 = 3
Point: (1, 3)
x = [2×4 + 3×(-1)]/(2+3) = (8-3)/5 = 1
y = [2×(-3) + 3×7]/5 = (-6+21)/5 = 3
Point: (1, 3)
22 marks
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Answer:
Let P(x₁,y₁) divide in ratio 1:2 → x₁ = [1×(-2)+2×4]/3 = (-2+8)/3 = 2, y₁ = [1×(-3)+2×(-1)]/3 = (-3-2)/3 = -5/3
Let Q(x₂,y₂) divide in ratio 2:1 → x₂ = [2×(-2)+1×4]/3 = (-4+4)/3 = 0, y₂ = [2×(-3)+1×(-1)]/3 = (-6-1)/3 = -7/3
Points: P(2, -5/3), Q(0, -7/3)
Let P(x₁,y₁) divide in ratio 1:2 → x₁ = [1×(-2)+2×4]/3 = (-2+8)/3 = 2, y₁ = [1×(-3)+2×(-1)]/3 = (-3-2)/3 = -5/3
Let Q(x₂,y₂) divide in ratio 2:1 → x₂ = [2×(-2)+1×4]/3 = (-4+4)/3 = 0, y₂ = [2×(-3)+1×(-1)]/3 = (-6-1)/3 = -7/3
Points: P(2, -5/3), Q(0, -7/3)
33 marks
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in Fig. 7.12. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the 8th line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Answer:
Coordinates: Green flag (2, 25), Red flag (8, 20).
Distance = √[(8-2)² + (20-25)²] = √[36+25] = √61 m.
Midpoint = ((2+8)/2, (25+20)/2) = (5, 22.5) → on 5th line at 22.5 m.
Coordinates: Green flag (2, 25), Red flag (8, 20).
Distance = √[(8-2)² + (20-25)²] = √[36+25] = √61 m.
Midpoint = ((2+8)/2, (25+20)/2) = (5, 22.5) → on 5th line at 22.5 m.
4-10Selected Solutions
Key results from remaining problems in Exercise 7.2:
Q4: Ratio 2:7 (x-coordinate gives 2:7)
Q5: Ratio 1:1 (midpoint)
Q6: Points: (-4, 6), (-3, 10), (2, 22) with ratio 1:2:1
Q7: Point (9/2, 6)
Q8: Point (8, -1)
Q9: Point (3, -2)
Q10: Ratio 3:4, Point (1, 2)
Q5: Ratio 1:1 (midpoint)
Q6: Points: (-4, 6), (-3, 10), (2, 22) with ratio 1:2:1
Q7: Point (9/2, 6)
Q8: Point (8, -1)
Q9: Point (3, -2)
Q10: Ratio 3:4, Point (1, 2)
Exercise 7.3 (Page 170) – Area of Triangle
12 marks each
Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)
(i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)
Answer:
(i) Area = ½|2(0+4) + (-1)(-4-3) + 2(3-0)| = ½|8 + 7 + 6| = 21/2 = 10.5 sq units
(ii) Area = ½|-5(-5-2) + 3(2+1) + 5(-1+5)| = ½|35 + 9 + 20| = 64/2 = 32 sq units
(i) Area = ½|2(0+4) + (-1)(-4-3) + 2(3-0)| = ½|8 + 7 + 6| = 21/2 = 10.5 sq units
(ii) Area = ½|-5(-5-2) + 3(2+1) + 5(-1+5)| = ½|35 + 9 + 20| = 64/2 = 32 sq units
23 marks
Find the area of triangle formed by points A(5,2), B(4,7), C(7,-4).
Answer:
Area = ½|5(7+4) + 4(-4-2) + 7(2-7)| = ½|55 - 24 - 35| = ½|-4| = 2 sq units
Area = ½|5(7+4) + 4(-4-2) + 7(2-7)| = ½|55 - 24 - 35| = ½|-4| = 2 sq units
33 marks
Find the value of k if points A(2,3), B(4,k), C(6,-3) are collinear.
Answer:
Area = 0 → ½|2(k+3) + 4(-3-3) + 6(3-k)| = 0
|2k+6 -24 + 18 -6k| = 0 → |-4k +0| = 0 → k = 0
Area = 0 → ½|2(k+3) + 4(-3-3) + 6(3-k)| = 0
|2k+6 -24 + 18 -6k| = 0 → |-4k +0| = 0 → k = 0
43 marks
Find the area of the quadrilateral whose vertices taken in order are (-4,-2), (-3,-5), (3,-2), (2,3).
Answer:
Divide into two triangles: ΔABC and ΔACD.
Area ΔABC = 21/2, ΔACD = 35/2 → Total = 28 sq units
Divide into two triangles: ΔABC and ΔACD.
Area ΔABC = 21/2, ΔACD = 35/2 → Total = 28 sq units
Exercise 7.4 (Optional – Page 171)
1-8Challenge Problems
Selected answers from optional exercise:
Q1: 2:9
Q2: (3, 4) and (3, -4)
Q3: Rectangle, area = 12 sq units
Q4: (6, 0), (0, 6), (0, 0)
Q5: Point (4, 0) with ratio 2:3
Q6: 4:1, internally
Q7: (3, 5), (1, 1)
Q8: (1, 3)
Q2: (3, 4) and (3, -4)
Q3: Rectangle, area = 12 sq units
Q4: (6, 0), (0, 6), (0, 0)
Q5: Point (4, 0) with ratio 2:3
Q6: 4:1, internally
Q7: (3, 5), (1, 1)
Q8: (1, 3)
Key Concepts Summary
Formula 1Distance Formula
What is the distance between two points P(x₁,y₁) and Q(x₂,y₂)?
Answer: PQ = √[(x₂-x₁)² + (y₂-y₁)²]
Formula 2Section Formula (Internal)
What are the coordinates of point dividing P(x₁,y₁) and Q(x₂,y₂) in ratio m:n internally?
Answer: ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))
Formula 3Area of Triangle
What is the area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃)?
Answer: Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
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