CG Board Class 10 Maths Chapter 6 Solutions: Co-Ordinate Geometry (निर्देशांक ज्यामिति) (2026-27)

by

CG Board (CGBSE / SCERT Chhattisgarh) Class 10 Mathematics Solutions | 2026-27 Edition

📚 Chapter Overview:

  • Board: Chhattisgarh Board of Secondary Education (CGBSE) / CG SCERT Raipur
  • Subject & Unit: Class 10 Mathematics – Unit 2: Co-Ordinate Geometry (2026-27 Edition)
  • Chapter 6: Co-Ordinate Geometry (निर्देशांक ज्यामिति)
  • Medium: Bilingual Step-by-Step Solutions (English Medium + हिंदी माध्यम)
📥 Download Mathematics 10 PDF (English)
📥 गणित कक्षा 10 पुस्तक PDF (Hindi)
📚 All 16 Chapters Maths Solutions Hub

🌐 Select Language / माध्यम चुनें:

Q1

Cartesian Plane and Origin


📝 Question:

Cartesian Plane and Origin*

💡 Solution:

In a plane, two mutually perpendicular lines XOX’ and YOY’ intersect at a central reference point O. The horizontal line XOX’ is called the x-axis, the vertical line YOY’ is called the y-axis, and the point of intersection O(0,0) is called the origin. This plane is called the Cartesian plane or coordinate plane.

Q2

Quadrants


📝 Question:

Quadrants*

💡 Solution:

The two perpendicular axes divide the Cartesian plane into 4 equal regions called quadrants (measured anti-clockwise from OX)

Q3

Coordinates Abscissa and Ordinate


📝 Question:

Coordinates: Abscissa and Ordinate*

💡 Solution:

For any point P(x, y)

  • Abscissa (x-coordinate): Perpendicular distance of the point from the y-axis, measured along the x-axis.
  • Ordinate (y-coordinate): Perpendicular distance of the point from the x-axis, measured along the y-axis.

Q4

Points on the Axes


📝 Question:

Points on the Axes*

💡 Solution:

Any point lying on the x-axis has y-coordinate zero, represented as (x, 0).

  • Any point lying on the y-axis has x-coordinate zero, represented as (0, y).

Q5

Distance Formula


📝 Question:

Distance Formula*

💡 Solution:

The distance between any two points P(x1, y1) and Q(x2, y2) in a Cartesian plane is given by

PQ = √((x2 – x1)2 + (y2 – y1)2)

  • Distance of a point P(x, y) from the origin O(0,0) is OP = √(x2 + y2).

Q6

Slope or Gradient of a Line


📝 Question:

Slope or Gradient of a Line*

💡 Solution:

Slope (m) represents the steepness of a line segment joining points A(x1, y1) and B(x2, y2)

Slope (m) = (y2 – y1)/(x2 – x1) = tan θ

where θ is the angle made by the line with the positive direction of the x-axis.

  • Horizontal Line (parallel to x-axis): y2 – y1 = 0 ⇒ Slope m = 0.
  • Vertical Line (parallel to y-axis): x2 – x1 = 0 ⇒ Slope m is undefined (∞).

Q7

Intercept and Equation of Line


📝 Question:

Intercept and Equation of Line*

💡 Solution:

The distance from the origin to the point where a line cuts the y-axis is called the y-intercept (c). The standard equation of a straight line with slope m and y-intercept c is

y = mx + c

Question 1

Coordinates of some points are given below Locate…


📝 Question:

Coordinates of some points are given below. Locate them on the Cartesian plane and write the quadrant in which the following points lie. 1. **(3, 4)

💡 Solution:

Here x = +3 > 0 and y = +4 > 0. Since both coordinates are positive (+, +), the point lies in the First Quadrant.

Here x = -5 < 0 and y = +6 > 0. With negative x and positive y (-, +), the point lies in the Second Quadrant.

Here x = -2 < 0 and y = -1 < 0. Since both coordinates are negative (-, -), the point lies in the Third Quadrant.

Here x = +2.5 > 0 and y = -7 < 0. With positive x and negative y (+, -), the point lies in the Fourth Quadrant.

Question 2

On the basis of the coordinates of the…


📝 Question:

On the basis of the coordinates of the following points, tell on which axis the points exist. 1. **(0, 5)

💡 Solution:

The x-coordinate is 0. Therefore, the point lies on the y-axis.

The y-coordinate is 0. Therefore, the point lies on the x-axis.

The y-coordinate is 0. Therefore, the point lies on the x-axis.

The x-coordinate is 0. Therefore, the point lies on the y-axis.

Question 3

Question 10


📝 Question:

**

💡 Solution:

Fill in the blanks

  1. Point P(-4, -7) lies in Third quadrant.
  2. The y-coordinate of any point on x-axis is Zero.
  3. On Cartesian plane, both the axes are mutually Perpendicular.
  4. The x-coordinate of any point on y-axis is Zero.
  5. Coordinates of origin are (0, 0).

Question 4

Observe the position of given points in Graph5…


📝 Question:

Observe the position of given points in Graph-5 and complete the task using the following instructions: Given points in Graph-5: A(1,1), B(3,6), C(5,5), D(3,3), E(-4,6), F(-3,1), G(-5,5), H(6,-1), P(3,-2), Q(-2,-2), R(-5,-5), S(5,-5). 1. Write the points whose x-coordinates are same: Points with x = 3: B, D, P Points with x = -5: G, R * Points with x = 5: C, S

💡 Solution:

B, D, P ; G, R ; C, S\**.

  1. Write the points whose y-coordinates are same:
  • Points with y = 6: B, E
  • Points with y = -2: P, Q
  • Points with y = 5: C, G
  • Points with y = -5: R, S

B, E ; P, Q ; C, G\.

  1. Write the points whose x- and y-coordinates are same:
  • Q(-2, -2), R(-5, -5), D(3, 3), C(5, 5)

Q, R, D, C.

Question 1

Question 12


📝 Question:

**

💡 Solution:

Calculate the distance of point P from Q and R.

  • Given coordinates from Graph-10: P(2, 0), Q(-1, 3), R(3, 2).
  • Distance PQ:

PQ = √((x2 – x1)2 + (y2 – y1)2)

PQ = √((-1 – 2)2 + (3 – 0)2) = √((-3)2 + 32) = √(9 + 9) = √18 = 3√2 units

  • Distance PR:

PR = √((3 – 2)2 + (2 – 0)2) = √(12 + 22) = √(1 + 4) = √5 units

  • Final Answer: PQ = 3√2 units, PR = √5 units .

Question 2

Question 13


📝 Question:

**

💡 Solution:

In Graph-10, find the values of AC, AB and BC.

  • Given coordinates from Graph-10: A(3, -2), B(6, -2), C(3, -4).
  • Finding AC:

Since x-coordinates are equal (x = 3)

AC = |y2 – y1| = |-4 – (-2)| = |-2| = 2 units

  • Finding AB:

Since y-coordinates are equal (y = -2)

AB = |x2 – x1| = |6 – 3| = 3 units

  • Finding BC:

BC = √((3 – 6)2 + (-4 – (-2))2) = √((-3)2 + (-2)2) = √(9 + 4) = √13 units

  • Final Answer: AC = 2, AB = 3, BC = √13 .

Question 3

Calculate the distance of point 3 4 from…


📝 Question:

Calculate the distance of point (3, 4) from the origin

💡 Solution:

Let point P(x, y) = (3, 4) and origin O(0, 0).

Using distance formula from origin OP = √(x2 + y2)

OP = √(32 + 42) = √(9 + 16) = √25 = 5 units

  • Final Answer: 5 units .

Question 4

Question 15


📝 Question:

**

💡 Solution:

If PA = PB and the coordinates of points A and B are (2, 0) and (-2, 4) respectively, and if P lies on the y-axis, then calculate the coordinates of point P.

(-2, 4) , P, y- , P )

  • Step 1: Since point P lies on the y-axis, let its coordinates be P(0, y).
  • Step 2: Express PA2 and PB2 using the distance formula

PA2 = (0 – 2)2 + (y – 0)2 = 4 + y2

PB2 = (0 – (-2))2 + (y – 4)2 = 22 + (y2 – 8y + 16) = y2 – 8y + 20

  • Step 3: Given PA = PB ⇒ PA2 = PB2

4 + y2 = y2 – 8y + 20

4 = -8y + 20

8y = 20 – 4

8y = 16 ⇒ y = 2

  • Final Answer: The coordinates of point P are P(0, 2) .

Question 5

Question 16


📝 Question:

**

💡 Solution:

Find the coordinates of the point which is on y-axis and which is equidistant from points (5, -2) and (3, 4).

(3, 4) )

  • Step 1: Let the required point on the y-axis be P(0, y).
  • Step 2: Let A(5, -2) and B(3, 4). Compute PA2 and PB2

PA2 = (0 – 5)2 + (y – (-2))2 = 25 + (y + 2)2 = y2 + 4y + 29

PB2 = (0 – 3)2 + (y – 4)2 = 9 + (y – 4)2 = y2 – 8y + 25

  • Step 3: Since P is equidistant from A and B, PA2 = PB2

y2 + 4y + 29 = y2 – 8y + 25

4y + 29 = -8y + 25

4y + 8y = 25 – 29

12y = -4 ⇒ y = -4/12 = -1/3

  • Final Answer: The required coordinates are (0, -1/3) .

Question 6

Question 17


📝 Question:

**

💡 Solution:

Find the relation between x and y such that point (x, y) is equidistant from the point (7, 1) and point (3, 5).

, (7, 1) (3, 5) )

  • Step 1: Let P(x, y), A(7, 1), and B(3, 5).
  • Step 2: Compute PA2 and PB2

PA2 = (x – 7)2 + (y – 1)2 = x2 – 14x + 49 + y2 – 2y + 1 = x2 + y2 – 14x – 2y + 50

PB2 = (x – 3)2 + (y – 5)2 = x2 – 6x + 9 + y2 – 10y + 25 = x2 + y2 – 6x – 10y + 34

  • Step 3: Set PA2 = PB2

x2 + y2 – 14x – 2y + 50 = x2 + y2 – 6x – 10y + 34

-14x – 2y + 50 = -6x – 10y + 34

-14x + 6x – 2y + 10y + 50 – 34 = 0

-8x + 8y + 16 = 0

Dividing the entire equation by -8

x – y – 2 = 0

  • Final Answer: x – y – 2 = 0 .

### *Exercise – 3

#### Question 1:

In Graph-19, find the slope or gradient of each interval.

Points from Graph-19: A(1, 5), B(3, 7), C(2, 4), D(5, 6), E(1, 3), F(2, 1), G(3, 2), H(6, 2), I(7, 5), J(7, 1).

  • Slope of AB:

m = (7 – 5)/(3 – 1) = 2/2 = 1

  • Slope of CD:

m = (6 – 4)/(5 – 2) = 2/3

  • Slope of EF:

m = (1 – 3)/(2 – 1) = -2/1 = -2

  • Slope of GH:

m = (2 – 2)/(6 – 3) = 0/3 = 0

  • Slope of IJ:

m = (1 – 5)/(7 – 7) = -4/0 = undefined

  • Final Answer: Slope of AB = 1, Slope of CD = 2/3, Slope of EF = -2, Slope of GH = 0, Slope of IJ = undefined .

Question 2

What is the slope of a line which…


📝 Question:

What is the slope of a line which is parallel to x-axis?

💡 Solution:

For a line parallel to the x-axis, y2 – y1 = 0. Thus, Slope m = (0)/(x2 – x1) = 0.

Question 3

A line whose slope is 56 passes through…


📝 Question:

A line whose slope is 5/6 passes through the point (7, 10). 1. Find the x-coordinate of that point whose y-coordinate is 15: Let the required point be (x, 15). Slope m = (y2 – y1)/(x2 – x1) 5/6 = (15 – 10)/(x – 7) 5/6 = (5)/(x – 7) ⇒ x – 7 = 6 ⇒ x = 13

💡 Solution:

x = 13** .

  1. What will be the value of x-coordinate when y-coordinate is -3?:

Let the point be (x, -3).

5/6 = (-3 – 10)/(x – 7)

5/6 = (-13)/(x – 7)

5(x – 7) = 6 × (-13)

5x – 35 = -78 ⇒ 5x = -43 ⇒ x = -43/5

x = -43/5 .

Question 4

A line passes through the points 3 7…


📝 Question:

A line passes through the points (3, 7) and (6, 8). Find the slope of the line

💡 Solution:

Given (x1, y1) = (3, 7) and (x2, y2) = (6, 8).

m = (y2 – y1)/(x2 – x1) = (8 – 7)/(6 – 3) = 1/3

  • Final Answer: 1/3 .

Question 5

Question 21


📝 Question:

**

💡 Solution:

Write the straight line 5x + 6y = 7 in the form y = mx + c, and find the slope of the line and its intercept on the y-axis.

  • Step 1: Convert 5x + 6y = 7 into slope-intercept form

6y = -5x + 7 ⇒ y = -5/6x + 7/6

  • Step 2: Comparing with y = mx + c
  • Slope (m) = -5/6
  • y-intercept (c) = 7/6
  • Final Answer: Form: y = -5/6x + 7/6, Slope = -5/6, Intercept = 7/6 .

Question 6

Find the equation of the straight line which…


📝 Question:

Find the equation of the straight line which cuts an intercept of 3 units on the y-axis and whose slope is 5/4

💡 Solution:

Given slope m = 5/4 and y-intercept c = 3.

Using formula y = mx + c

y = 5/4x + 3

Multiplying by 4

4y = 5x + 12 ⇒ 5x – 4y + 12 = 0

  • Final Answer: 5x – 4y + 12 = 0 .

Question 7

What is the slope of a line parallel…


📝 Question:

What is the slope of a line parallel to the y-axis?

💡 Solution:

For a line parallel to the y-axis, x2 – x1 = 0. Division by zero is undefined.

Question 8

Find the equation of that line which cuts…


📝 Question:

Find the equation of that line which cuts an intercept of 6 units from the y-axis and whose slope is -5/3

💡 Solution:

Given slope m = -5/3 and y-intercept c = 6.

Using y = mx + c

y = -5/3x + 6

Multiplying by 3

3y = -5x + 18 ⇒ 5x + 3y – 18 = 0

  • Final Answer: 5x + 3y – 18 = 0 .

Question 9

Find the equation of the line which passes…


📝 Question:

Find the equation of the line which passes through the point (6, 0) and whose slope is 7/3

💡 Solution:

Given point (x1, y1) = (6, 0) and slope m = 7/3.

Using point-slope formula y – y1 = m(x – x1)

y – 0 = 7/3(x – 6)

3y = 7(x – 6)

3y = 7x – 42 ⇒ 7x – 3y – 42 = 0

  • Final Answer: 7x – 3y – 42 = 0 .

Question 10

Find the slope of the line which passes…


📝 Question:

Find the slope of the line which passes through the origin and also passes through the point (2, 3)

💡 Solution:

Points are Origin (x1, y1) = (0, 0) and (x2, y2) = (2, 3).

m = (y2 – y1)/(x2 – x1) = (3 – 0)/(2 – 0) = 3/2

  • Final Answer: 3/2 .

← Chapter 5 Solutions
📚 Class 10 Maths All Chapters Hub
Chapter 7 Solutions →

Leave a Reply

📑 Chapter Navigation Directory

CG Board Class 10 Mathematics Solutions
CG Board Solutions Directory