NCERT Solutions for Class 10 Mathematics Chapter 7
NCERT solutions for Class 10 Mathematics Chapter Coordinate Geometry – complete answers & explanations
Coordinate Geometry is an important chapter in Class 10 Mathematics that helps students understand how algebra and geometry are connected through coordinates. In this chapter, students learn how to calculate distances between points, check collinearity, find midpoints, and divide line segments in given ratios. These concepts are very useful in solving real-life problems related to maps, positioning, and measurements. These NCERT solutions provide clear and reliable answers that help students follow the correct approach step by step. With regular practice, students can improve their problem-solving skills and accuracy in calculations. Download the worksheet and practice alongside solutions for better clarity. Book a free trial now to get expert guidance. This chapter also builds a strong base for higher-level mathematics topics.
What this NCERT chapter covers?
1. Understanding the coordinate plane and
plotting points using ordered pairs.
2. Learning the distance formula to find the distance between two points.
3. Checking whether three points are collinear using distance relationships.
4. Identifying different types of triangles using coordinate geometry.
5. Solving problems related to quadrilaterals like square, rhombus, and parallelogram.
6. Finding unknown coordinates using the distance formula.
7. Understanding the section formula to divide a line segment in a given ratio.
8. Learning about midpoint formula and its applications.
9. Solving problems involving trisection and internal division of line segments.
10. Applying coordinate geometry concepts to calculate area and distances in figures.
How to use these NCERT solutions?
1. Attempt each question from the
worksheet on your own before checking the solutions.
2. Use the answers to verify your steps and correct mistakes if needed.
3. Carefully follow the formulas used in each solution.
4. Practice similar questions multiple times to gain confidence.
5. Focus on understanding the calculation steps instead of memorizing answers.
6. Follow the exact order of questions as given in the worksheet.
7. Parents and teachers can guide students by explaining each step clearly.
8. Revise formulas regularly to improve speed and accuracy.
mportant tips & tricks for students
1. Always write the distance formula correctly before substituting values.
2. Double-check signs while calculating coordinates to avoid mistakes.
3. Practice plotting points correctly on the coordinate plane.
4. Learn and revise section and midpoint formulas regularly.
5. Show all calculation steps clearly to score full marks.
6. Avoid skipping steps in long calculations.
7. Understand how to identify shapes using distances and slopes.
8. Practice more problems to improve speed and accuracy.
NCERT solutions – complete answer key
EXERCISE 7.1
1
(i) √8
(ii) √32
(iii) 2√(a² + b²)
2
Distance = 39 km
3
Points: (1, 5), (2, 3), (–2, –11)
AB = √[(2–1)² + (3–5)²] = √[1 + 4] = √5
BC = √[(–2–2)² + (–11–3)²] = √[16 + 196] = √212
AC = √[(–2–1)² + (–11–5)²] = √[9 + 256] = √265
Since AB + BC ≠ AC, AB + AC ≠ BC, BC + AC ≠ AB
Points are NOT collinear
4
Points: (5, –2), (6, 4), (7, –2)
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
5
Using distance formula, calculate all sides:
If AB = BC = CD = DA and diagonals equal → square
Thus, Champa is correct → ABCD is a square
6
(i)
All sides equal → Rhombus
(ii)
Sides not equal, diagonals unequal → No special
quadrilateral
(iii)
Opposite sides equal → Parallelogram
7
Let point be (x, 0)
Distance from (2, –5) = distance from (–2, 9)
(x–2)² + (0+5)² = (x+2)² + (0–9)²
(x–2)² + 25 = (x+2)² + 81
x² – 4x + 4 + 25 = x² + 4x + 4 + 81
–4x + 29 = 4x + 85
–8x = 56
x = –7
Point = (–7, 0)
8
Distance formula:
√[(10–2)² + (y+3)²] = 10
64 + (y+3)² = 100
(y+3)² = 36
y+3 = ±6
y = 3 or y = –9
9
Q(0,1) equidistant from P(5,–3) and R(x,6)
QP² = QR²
(0–5)² + (1+3)² = (0–x)² + (1–6)²
25 + 16 = x² + 25
41 = x² + 25
x² = 16
x = ±4
QR = √[(0–4)² + (1–6)²] = √(16+25) = √41
PR = √[(5–4)² + (–3–6)²] = √(1+81) = √82
10
Let P(x,y) equidistant:
(x–3)² + (y–6)² = (x+3)² + (y–4)²
Expand:
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
EXERCISE 7.2
1
Using section formula:
x = [2(4) + 3(–1)] / 5 = (8 – 3)/5 = 1
y = [2(–3) + 3(7)] / 5 = (–6 + 21)/5 = 3
Point = (1, 3)
2
Trisection points:
First point (ratio 1:2): (2, –5/3)
Second point (ratio 2:1): (0, –7/3)
3
Use section formula to find both flag positions.
Then apply distance formula for distance.
Midpoint = ((x1+x2)/2, (y1+y2)/2)
4
Let ratio = m1 : m2
Using section formula → ratio = 2 : 1
5
Let point = (x,0)
Using section formula:
0 = [m1(5) + m2(–5)] / (m1 + m2)
m1 = m2
Ratio = 1 : 1
Point = midpoint = (–1.5, 0)
6
Midpoints equal:
(1+ x)/2 = (4+3)/2 → x = 2
(2+6)/2 = (y+5)/2 → y = 3
7
Midpoint = (2, –3)
( x + 1 ) / 2 = 2 → x = 3
( y + 4 ) / 2 = –3 → y = –10
Point = (3, –10)
8
AP = (3/7) AB
Using section formula:
P = [3(2) + 4(–2)]/7 , [3(–4) + 4(–2)]/7
= (–2/7, –20/7)
9
Divide into 4 equal parts:
Points:
(–1, 3.5), (0, 5), (1, 6.5)
10
Diagonals:
d1 = √[(3+1)² + (0–4)²] = √(16+16)=√32
d2 = √[(4+2)² + (5+1)²] = √(36+36)=√72
Area = ½ × d1 × d2
= ½ × √32 × √72
= ½ × √2304
= ½ × 48 = 24 sq units
Why NCERT solutions help students?
NCERT solutions help students build strong conceptual understanding and improve accuracy in solving coordinate-based problems. They also help in exam preparation by providing the correct approach and step-by-step solutions, making it easier for students to score well and gain confidence.
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