Is RS Aggarwal enough for class 10 maths?

RS Aggarwal alone can certainly help you score well in class 10 board exams and achieve a score of 95 marks.

Is it easy to score 100 in maths class 10?

While scoring a perfect 100 in the Class 10 Maths Board Exam may seem like a daunting task, with the right strategy and dedication, it's achievable.

Is class 10 maths tough or easy?

The difficulty level of the CBSE Math Class 10 questions was easy to moderate.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 Coordinate Geometry

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3: The Physics Wallah academic team has produced a comprehensive answer for Chapter 16: Coordinate Geometry in the RS Aggarwal class 10 textbook. The RS Aggarwal class 10 solution for chapter 16 Coordinate Geometry Exercise-16C is uploaded for reference only; do not copy the solutions.

Before going through the solution of Chapter 16 Coordinate Geometry Exercise-16C, one must have a clear understanding of Chapter 16 Coordinate Geometry. Read the theory of chapter 16 Coordinate Geometry and then try to solve all numerical of exercise-16C. Complete the NCERT exercise questions and utilize them as a guide. You need assistance to get through the Class 10 Math questions in Physics Wallah NCERT solutions. Class 10 Math NCERT solutions were uploaded by Physics Wallah.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 Coordinate Geometry Overview

Chapter 16 of RS Aggarwal's Class 10 Maths textbook focuses on coordinate geometry, a branch of mathematics that deals with the study of geometric figures using the coordinate system. Exercise 16.3 specifically delves into the various concepts and applications of the coordinate plane, including the calculation of distances between points, the determination of the coordinates of a point dividing a line segment in a given ratio, and the finding of areas of triangles formed by given points. By working through Exercise 16.3, students enhance their understanding of the spatial relationships between points in a plane and develop problem-solving skills essential for higher mathematics. The questions are designed to solidify the theoretical concepts of coordinate geometry through practical application, ensuring a comprehensive grasp of the topic.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 provides a comprehensive learning experience that equips students with the necessary knowledge and skills to tackle questions confidently. Here we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 for the ease of students so that they can prepare better for their upcoming exams –

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 (Ex 16C)

Below we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 for the ease of the students –

Question

Find the area of $△ A B C$ whose vertices are :

(i) A(1, 2), B (-2, 3) and C(-3, -4) (ii) A (-5,7),B(-4, -5) and C(4,5) (iii) A(3, 8), B(-4, 2) and C(5, -1) (iv) A (10, -6),B (2,5) and C(-1, 3)

Solution

(i) A (1, 2), B (-2, 3) and C (-3, -4) are the vertices of ΔABC. Then

(x_{1} = 1, y_{1} = 2), (x_{2} = - 2, y_{2} = 3), (x_{3} = - 3, y_{3} = - 4)

Area of triangle ABC

= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [1 (3 - (- 4)) + (- 2) (- 4 - 2) + (- 3) (2 - 3)] = \frac{1}{2} [1 (3 + 4) - 2 (- 6) - 3 (- 1)] = \frac{1}{2} [7 + 12 + 3] = \frac{1}{2} [22] = 11 s q . u n i t s

(ii) A(-5, 7), B(-4, -5) and C(4, 5) A (-5, 7), B (-4, -5) and C (4, 5) are the vertices of ΔABC. Then

(x_{1} = - 5, y_{1} = 7), (x_{2} = - 4, y_{2} = - 5), (x_{3} = 4, y_{3} = 5)

Area of triangle ABC

= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [- 5 (- 5 - 5) + (- 4) (5 - 7) + 4 (7 - (- 5))] = \frac{1}{2} [- 5 (- 10) - 4 (- 2) + 4 (12)] = \frac{1}{2} [50 + 8 + 48] = \frac{1}{2} [106] = 53 s q . u n i t s

(iii) A(3, 8), B(-4, 2) and C(5, -1) A (3, 8), B (-4, 2) and C (5, -1) are the vertices of ΔABC. Then

(x_{1} = 3, y_{1} = 8), (x_{2} = - 4, y_{2} = 2), (x_{3} = 5, y_{3} = - 1)

Area of triangle ABC

= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [3 (2 - (- 1)) + (- 4) (- 1 - 8) + 5 (8 - 2))] = \frac{1}{2} [3 (2 + 1) - 4 (- 9) + 5 (6)] = \frac{1}{2} [9 + 36 + 30] = \frac{1}{2} [75] = 37.5 s q . u n i t s

(iv) A(10, -6), B(2, 5) and C(-1, 3) A (10, -6), B (2, 5) and C (-1, -3) are the vertices of ΔABC. Then

(x_{1} = 10, y_{1} = - 6), (x_{2} = 2, y_{2} = 5), (x_{3} = - 1, y_{3} = 3)

Area of triangle ABC

= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [10 (5 - 3) + (2) (3 - (- 6)) + (- 1) (- 6 - 5)] = \frac{1}{2} [10 (2) + 2 (9) - 1 (11)] = \frac{1}{2} [20 + 18 + 11] = \frac{1}{2} [49] = 24.5 s q . u n i t s

Question

Find the area of quadrilateral ABCD whose vertices are A(3, -1), B(9, -5), C(14, 0) and D(9, 19).

Solution

By joining A and C, we get two triangles ABC and ACD Let A (3, -1), B (9, -5) and C (14, 0) and D (9, 19)

(x_{1} = 3, y_{1} = - 1), (x_{2} = 9, y_{2} = - 5), (x_{3} = 14, y_{3} = 0), (x_{4} = 9, y_{4} = 19)

Then Area of triangle ABC

= \frac{1}{2} (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))

= \frac{1}{2} \times (3 (- 5 - 0) + (9) (0 + 1) + (14) (- 1 + 5))

= \frac{1}{2} \times (3 (- 5) + 9 (1) + 14 (4))

= \frac{1}{2} \times (- 15 + 9 + 56)

= \frac{1}{2} \times (50)

= 25 sq.units Area of triangle ACD

\frac{1}{2} \times (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))

= \frac{1}{2} \times (3 (0 - 19) + (14) (19 + (1)) + (9) (- 1 - 0)))

= \frac{1}{2} \times (3 (- 19) + 14 (20) + 9 (- 1))

= \frac{1}{2} \times (- 57 + 280 - 9)

= \frac{1}{2} \times (214)

= 107 sq.units So, the area of the quadrilateral is 25 + 107 = 132 sq. units

Question

Find the area of quadrilateral PQRS whose vertices are P(-5, -3), Q(-4,-6), R(2,-3) and S(1,2).

Solution

By joining P and R, we get two triangles PQR and PRS Let P (-5, -3), Q (- 4, – 6) and R (2, -3) and S (1, 2)

(x_{1} = - 5, y_{1} = - 3), (x_{2} = - 4, y_{2} = - 6), (x_{3} = 2, y_{3} = - 3), (x_{4} = 1, y_{4} = 2)

Then Area of triangle PQR

= \frac{1}{2} (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))

= \frac{1}{2} \times (- 5 (- 6 + 3) - 4 (- 3 + 3) + 2 (- 3 + 6))

= \frac{1}{2} \times (- 5 (- 3) - 4 (0) + 2 (3))

= \frac{1}{2} \times (15 - 0 + 6)

= \frac{1}{2} \times (21)

= 10.5 sq.units Area of triangle PRS

\frac{1}{2} \times (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))

= \frac{1}{2} \times (- 5 (- 3 - 2) + 2 (2 + 3) + 1 (- 3 + 3))

= \frac{1}{2} \times (- 5 (- 5) + 2 (5) + 1 (0))

= \frac{1}{2} \times (25 + 10 + 0)

= \frac{1}{2} \times (35)

= 17.5 sq.units So, the area of the quadrilateral is 10.5 + 17.5 = 28 sq. units

Question

If A(-7, 5), B(-6, -7), C (-3,-8) and D(2,3) are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.

Solution

By joining A and C, we get two triangles ABC and ACD Let A (-7,5), B (-6, -7) and C (-3, -8) and D (2, 3)

(x_{1} = - 7, y_{1} = 5), (x_{2} = - 6, y_{2} = - 7), (x_{3} = - 3, y_{3} = - 8), (x_{4} = 2, y_{4} = 3)

Then Area of triangle ABC

= \frac{1}{2} (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))

= \frac{1}{2} \times (- 7 (- 7 + 8) - 6 (- 8 - 5) - 3 (5 + 7))

= \frac{1}{2} \times (- 7 (1) - 6 (- 13) - 3 (12))

= \frac{1}{2} \times (- 7 + 78 - 36)

= \frac{1}{2} \times (35)

= 17.5 sq.units Area of triangle ACD

= \frac{1}{2} \times (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))

= \frac{1}{2} \times (- 7 (- 8 - 3) - 3 (3 - 5) + 2 (5 + 8))

= \frac{1}{2} \times (- 7 (- 11) - 3 (- 2) + 2 (13))

= \frac{1}{2} \times (74 + 6 + 26)

= \frac{1}{2} \times (106)

= 53 sq.units So, the area of the quadrilateral is 53+ 17.5 = 70.5 sq. units

Question

Find the area of $△ A B C$ with A(1, -4) and midpoints of sides through A being (2, -1) and (0, -1)

Solution

Let

x_{2}, y_{2}

and

x_{3}, y_{3}

be the co-ordinates of B and C respectively. Since, the co-ordinates of A (1, -4) hence let us name midpoint of AB be D =

x_{2}, y_{2}

and midpoint of AC be E=

x_{3}, y_{3}

. Now D (

2, - 1)

\frac{(1 + x_{2})}{2}

\frac{(1 + y_{2})}{2}

\to

2 =

\frac{(1 + x_{2})}{2}

\to 4 = 1 + x_{2}

\to 3 = x_{2}

\to x_{2} = 3

Again

- 1 = \frac{(- 4 + y_{2})}{2}

- 4 + y_{2} = - 2

\to y_{2} = - 2 + 4 = 2

Similarly Now E

0, - 1

\to 0 = \frac{(1 + x_{3})}{2}

\to 0 = - 1 + x_{3}

\to - 1 = x_{3}

\to x_{3} = - 1

From E

0, - 1

-1=

\frac{(- 4 + y_{3})}{2}

\to y_{3} = 2

Let A

{x_{1}, y_{1}) = A (1, - 4)

Let B

{x_{2}, y_{2}) = B (3, 2)

Let C

{x_{3}, y_{3}) = C (- 1, 2)

Now Area of

△ A B C

\frac{1}{2} [(x_{1} (y_{2} - y_{3})] + [x_{2} (y_{3} - y_{1}] + [x_{3} (y_{1} - y_{2})]

\frac{1}{2} \times [1 (2 - 2) + 3 (2 + 4) - 1 (- 4 - 2)]

\frac{1}{2} \times [1 (0) + 3 (6) - 1 (- 6)]

\frac{1}{2} \times [0 + 18 + 6]

\frac{1}{2} \times [24]

= 12 sq.unit Thus, Area of

△ A B C

= 12 sq.units

Question

(i) If the vertices of $△ A B C$ be A(1, -3), B(4, p) and C(-9, 7) and its area is 15 square units, find the values of p.

(ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is $(\frac{7}{2}, y)$ , find the value of y.

Solution

(i) Let

A (x_{1}, y_{1}) = A (1, - 3)

B (x_{2}, y_{2}) = B (4, p)

and

C (x_{3}, y_{3}) = C (- 9, 7)

. Now

A r e a (Δ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \Rightarrow 15 = \frac{1}{2} [1 (p - 7) + 4 (7 + 3) - 9 (- 3 - p)] \Rightarrow 15 = \frac{1}{2} [10 p + 60] \Rightarrow | 10 p + 60 | = 30 \Rightarrow 10 p + 60 = - 30 o r 30 \Rightarrow 10 p = - 90 o r - 30 \Rightarrow p = - 9 o r - 3

Hence, p = -9 or p = -3 (ii) Let

A (x_{1}, y_{1}) = A ((2, 1))

B (x_{2}, y_{2}) = B (3, - 2)

and

C (x_{3}, y_{3}) = C (\frac{7}{2}, y)

. Now

A r e a (Δ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \Rightarrow 5 = \frac{1}{2} [2 (- 2 - y) + 3 (y - 1) + \frac{7}{2} (1 + 2)] \Rightarrow 5 = \frac{1}{2} [- 4 - 2 y + 3 y - 3 + \frac{21}{2}] \Rightarrow 10 = [y - 7 \frac{21}{2}] \Rightarrow 10 = | \frac{2 y - 14 + 21}{2} | \Rightarrow 10 = | \frac{2 y + 7}{2} | \Rightarrow 20 = | 2 y + 7 | \Rightarrow 20 = 2 y + 7 o r - 20 = 2 y + 7 \Rightarrow 2 y = 20 - 7 o r 2 y = - 20 - 7 \Rightarrow 2 y = 13 o r 2 y = - 27 \Rightarrow y = \frac{13}{2} o r y = \frac{- 27}{2}

Question

The are of $△ A B C$ with vertices A(a, 0), O(0, 0) and B(0, b) in square units is

(a) ab (b) $\frac{1}{2} a b$ (c) $\frac{1}{2} a^{2} b^{2}$ (d) $\frac{1}{2} b^{2}$

Solution

A (a , 0), O (0 , 0) and B (0 , b) are the vertices of ΔABC. Then

(x_{1} = a, y_{1} = 0), (x_{2} = 0, y_{2} = 0), (x_{3} = 0, y_{3} = b)

Area of triangle ABC =

\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]

\frac{1}{2} [a (0 - b) + 0 (b - 0) + 0 (0 - 0)]

\frac{1}{2} [a (- b)]

| \frac{1}{2} [a (- b)] |

\frac{1}{2} a b

answer :- (b)

\frac{1}{2} a b

Question

If P $(\frac{a}{2}, 4)$ is the midpoint of the line segment joining the points A (-6, 5) and B(-2, 3) then the value of a is

(a) -8 (b) 3 (c) -4 (d) 4

Solution

$i f space t h e space p o i n t space open parentheses x comma y close parentheses space d i v i d e s space t h e space j o i n space o f space t h e space p o i n t s space open parentheses x subscript 1 comma y subscript 1 close parentheses space a n d open parentheses x subscript 2 comma y subscript 2 close parentheses space i n space t h e space r a t i o space m colon n t h e n open parentheses x comma y close parentheses equals open parentheses fraction numerator x subscript 1 n plus x subscript 2 m over denominator m plus n end fraction comma fraction numerator y subscript 1 n plus y subscript 2 m over denominator m plus n end fraction close parentheses i n space t h i s space q u e s t i o n open parentheses a over 2 comma 4 space close parentheses space i n space m i d space p o i n t space o f space l i n e space A B s o m equals n equals 1 x equals a over 2 comma y equals 4 comma x subscript 1 equals negative 6 space comma x subscript 2 equals negative 2 space comma y subscript 1 equals 5 comma y subscript 2 equals 3 t h e n x equals fraction numerator x subscript 1 plus x subscript 2 over denominator 2 end fraction equals a over 2 a over 2 equals fraction numerator negative 6 plus negative 2 over denominator 2 end fraction a equals negative 8 s o space o p t i o n space a space i s space c o r r e c t$

Question

Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Solution:

Let the coordinates of point A be ( x , y ). Mid-point of AB is (2, – 3), which is the centre of the circle. Coordinate of B = (1, 4) (2, -3) =((x+1)/2 , (y+4)/2) (x+1)/2 = 2 and (y+4)/2 = -3 x + 1 = 4 and y + 4 = -6 x = 3 and y = -10 The coordinates of A(3,-10).

Question

If the point A (x, 2) is equidistant from the points B(8, -2) and C(2, -2) find the value of x. Also, find the length of AB.

Solution

As per the question, we have AB=AC

\sqrt{x - 8)^{2} + {(2 + 2)}^{2}} = \sqrt{{(x - 2)}^{2} + {(2 + 2)}^{2}}

\sqrt{x^{2} + 64 - 16 \times x + 16} = \sqrt{x^{2} + 4 - 4 \times x + 16}

\sqrt{x^{2} - 16 \times x + 80} = \sqrt{x^{2} - 4 \times x + 20}

squaring both sides we get

x^{2} - 16 x + 80 = x^{2} - 4 x + 20

80-20=-4x+16x 80-20=-4x+16x x=5 AB=

\sqrt{{(x - 8)}^{2} + {(2 + 2)}^{2}}

\sqrt{{(5 - 8)}^{2} + {(4)}^{2}}

\sqrt{9 + 16} = \sqrt{25}

= \pm5 \) hence AB=5

Question

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7).

Solution

Consider line 2x + y – 4 = 0 divides line AB joined by the two points A(2, -2) and B(3, 7) in k:1 ratio. Coordinates of point of division can be given as follows: x = (2 + 3k)/(k + 1) and y = (-2 + 7k)/(k + 1) Substituting the values of x and y given equation, i.e. 2x + y – 4 = 0, we have 2{(2 + 3k)/(k + 1)} + {(-2 + 7k)/(k + 1)} – 4 = 0 (4 + 6k)/(k + 1) + (-2 + 7k)/(k + 1) = 4 4 + 6k – 2 + 7k = 4(k+1) -2 + 9k = 0 Or k = 2/9 Hence, the ratio is 2:9.

Question

Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).

Solution

Let A = (6, -6), B = (3, -7), and C = (3, 3) are the points on a circle. If O is the centre, then OA = OB = OC (radii are equal) If O = (x, y), then OA = √[(x – 6) ² + (y + 6) ² ] OB = √[(x – 3) ² + (y + 7) ² ] OC = √[(x – 3) ² + (y – 3) ² ] Choose: OA = OB, we have After simplifying above, we get -6x = 2y – 14 ….(1) Similarly, OB = OC (x – 3) ² + (y + 7) ² = (x – 3) ² + (y – 3) ² (y + 7) ² = (y – 3) ² y ² + 14y + 49 = y ² – 6y + 9 20y =-40 or y = -2 Substituting the value of y in equation (1), we get -6x = 2y – 14 -6x = -4 – 14 = -18 x = 3 Hence, the centre of the circle is located at point (3,-2).

Question

Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).

Solution:

Consider the ratio in which the line segment joining (-3, 10) and (6, -8) is divided by point (-1, 6) be k :1. Therefore, -1 = ( 6 k -3)/( k +1) – k – 1 = 6 k -3 7 k = 2 k = 2/7 Therefore, the required ratio is 2: 7.

Question

In each of the following, find the value of ‘k’, for which the points are collinear.

(i) (7, -2), (5, 1), (3, -k)

(ii) (8, 1), (k, -4), (2, -5)

Solution:

(i) For collinear points, the area of triangle formed by them is always zero. Let points (7, -2), (5, 1), and (3, k) are vertices of a triangle. Area of triangle = 1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) – 1}] = 0 7 – 7k + 5k +10 -9 = 0 -2k + 8 = 0 k = 4 (ii) For collinear points, the area of triangle formed by them is zero. Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0 1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0 8 – 6k + 10 = 0 6k = 18 k = 3

Question

Find the area of the triangle whose vertices are:

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

Solution:

Area of a triangle formula = 1/2 × [x ₁ (y ₂ – y ₃ ) + x ₂ (y ₃ – y ₁ ) + x ₃ (y ₁ – y ₂ )] (i) Here, x ₁ = 2, x ₂ = -1, x ₃ = 2, y ₁ = 3, y ₂ = 0 and y ₃ = -4 Substitute all the values in the above formula, we get Area of triangle = 1/2 [2 {0- (-4)} + (-1) {(-4) – (3)} + 2 (3 – 0)] = 1/2 {8 + 7 + 6} = 21/2 So, the area of the triangle is 21/2 square units. (ii) Here, x ₁ = -5, x ₂ = 3, x ₃ = 5, y ₁ = -1, y ₂ = -5 and y ₃ = 2 Area of the triangle = 1/2 [-5 { (-5)- (2)} + 3(2-(-1)) + 5{-1 – (-5)}] = 1/2{35 + 9 + 20} = 32 Therefore, the area of the triangle is 32 square units.

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3

The RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 offer numerous benefits to students studying coordinate geometry:

Conceptual Clarity : The solutions provide step-by-step explanations for each problem, helping students understand the fundamental concepts of coordinate geometry such as the distance formula, section formula, and area of triangles.

Enhanced Problem-Solving Skills : By working through various types of problems, students can develop strong problem-solving skills. The solutions illustrate different methods to approach and solve coordinate geometry problems, making students adept at tackling similar questions in exams.

Time Management : With clear and concise solutions, students can learn to solve problems more efficiently, improving their time management skills during exams.

Accuracy : The solutions ensure that students follow the correct procedures and calculations, reducing errors and increasing accuracy in their work.

Confidence Building : Regular practice with accurate solutions builds confidence in students, as they can verify their answers and understand their mistakes, leading to improved performance in tests and exams.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 Coordinate Geometry

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 FAQs

Is RS Aggarwal enough for class 10 maths?

Who is the father of coordinate geometry class 10?

Is it easy to score 100 in maths class 10?

Is class 10 maths tough or easy?

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 FAQs

Is RS Aggarwal enough for class 10 maths?

Who is the father of coordinate geometry class 10?

Is it easy to score 100 in maths class 10?

Is class 10 maths tough or easy?

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 Coordinate Geometry

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 Coordinate Geometry Overview

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3 (Ex 16C)

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.3