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RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2

Here, we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2. Students can view these RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2 before exams for better understanding.
authorImageAnanya Gupta25 Jul, 2024
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RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2: RS Aggarwal Solutions for Class 10 Maths Chapter 16, Exercise 16.2 provide detailed answers and explanations for problems related to Coordinate Geometry. This exercise focuses on applying concepts such as the distance formula, midpoint theorem, and section formula to solve a variety of coordinate geometry problems.

By working through this exercise, students can enhance their problem-solving skills, solidify their understanding of key concepts and prepare effectively for their exams.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2 Overview

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2 are created by experts from Physics Wallah. This exercise helps students understand Coordinate Geometry by solving problems with clear, step-by-step solutions. The experts explain how to use important formulas like the distance formula, midpoint theorem, and section formula. These solutions make it easier for students to learn and apply these concepts, which is very useful for preparing for exams.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2 PDF

The PDF link for RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2 is available below. By downloading this PDF, students can access expert-prepared solutions and improve their problem-solving skills making it a valuable resource for their exam preparation.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2

Here we have provided the RS Aggarwal Solutions for Class 10 Maths Chapter 16 Coordinate Geometry Exercise 16.2 to help students prepare better for their exams. These solutions are designed to make it easier for students to understand and practice key concepts in Coordinate Geometry ensuring they are well-prepared and confident for their upcoming tests.

Q. If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of the point P such that AP = 3 7 AB, where P lies on the line segment AB.

Solution:
The coordinates of the points A and B are (-2,-2) and (2,-4) respectively where A P = 3 7 A B and P lies on the line segment AB. So A P + B P = A B ⇒ A P + B P = 7 A P 3 A B ( ∵ A P = 3 7 A B ) ⇒ B P = 7 A P 3 − A P = 4 A P 3 ⇒ A P B P = 3 4 Let (x,y) be the coordinates of P which divides AB in the ratio 3:4 internally. Then Therefore, ( x 1 = − 2 , y 1 = − 2 ) and ( x 2 = 2 , y 2 = − 4 ) Also, m = 3 and n = 4 Let the required point be P(x,y) By section formula, we get x = ( m x 2 + n x 1 m + n , y = m y 2 + n y 1 m + n ) ⇒ x = ( 3 × 2 ) + ( 4 × − 2 ) 3 + 4 ⇒ x = 6 − 8 7 ⇒ x = − 2 7 ⇒ y = ( 3 × − 4 ) + ( 4 × − 2 ) 3 + 4 ⇒ y = − 12 − 8 7 ⇒ y = − 20 7 Hence, the coordinates of the point P are ( − 2 7 , − 20 7 )

Q. Point A lies on the line segment PQ joining P(6, -6) and Q(-4, -1) in such a way that P A P Q = 2 5 . If the point A also lies on the line 3x+k(y+1) = 0, find the value of k.

Solution:
Point A divides PQ in the ratio of 2:3 internally . so coordinates of A are ; A= ( 2 ( − 4 ) + 3 ( 6 ) ( 2 + 3 ) , 2 ( − 1 ) + 3 ( − 6 ) ( 2 + 3 ) ) A=(2,-4) if A lies on 3x+k(y+1) = 0, Then 3(2)+k(-4+1)=0 6+k(-3)=0 k=2

Q. Points P, Q, R and S divide the line segment joining the points A (1, 2) and B(6, 7) into five equal parts. Find the coordinates of the points P, Q and R.

Solution:
P divides AB in the ratio is 1:4 Coordinates of P using section formula, using Section Formula given by. 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 Here m = 1 and n = 4, X subscript 1 = 6, x subscript 2 = 1, y subscript 1 = 7, y subscript 2 = 2 x equals fraction numerator 1 cross times 6 plus 4 cross times 1 over denominator 1 plus 4 end fraction equals 2 space comma y equals fraction numerator 1 cross times 7 plus 4 cross times 2 over denominator 1 plus 4 end fraction equals 3 Coordinates of P (2,3). Q divides AB in the ratio is 2:3 Coordinates of Q x equals fraction numerator 2 cross times 6 plus 3 cross times 1 over denominator 2 plus 3 end fraction equals 3 comma y equals fraction numerator 2 cross times 7 plus 3 cross times 2 over denominator 2 plus 3 end fraction equals 4 Coordinates of Q are (3,4) R divides AB in the ratio is 3:2 Coordinates of R are x equals fraction numerator 3 cross times 6 plus 2 cross times 1 over denominator 3 plus 2 end fraction equals 4 comma y equals fraction numerator 3 cross times 7 plus 2 cross times 2 over denominator 3 plus 2 end fraction equals 5 Coordinates of R are (4,5)

Q. The line segment joining the points A (3, -4) and B (1, 2) is trisected at the points P(p, -2) and Q ( 5 3 , q ) . Find the values of p and q.

Solution:
We know that a ratio m:n divides with coordinates ∴ P ( x , y ) = ( m x 2 + n x 1 m + n , m y 2 + n y 1 m + n ) Here trisection points are P(p, -2) and Q ( 5 3 ,q) and points are A (3, -4) and B (1, 2). Trisection can occur in either 1:2 or 2:1 ratio. But we don't know which point (P or Q) trisects in which ratio. lets think P divides in m:n ratio and so y = m y 2 + n y 1 m + n − 2 = 2 m + ( − 4 ) n m + n ) − 2 ( m + n ) = 2 m − 4 n − 2 m − 2 n = 2 m − 4 n − 4 m = − 2 n m n = 2 4 = 1 2 So P divides in the ratio 1:2 now p = m x 2 + n x 1 m + n = 1 × 1 + 2 × 3 3 = 1 + 6 3 = 7 3 now for Q, 5 3 = m x 2 + n x 1 m + n 5 3 = m + 3 n m + n 5 m + 5 n = 3 m + 9 n 2 m = 4 n f r a c m n = 4 2 = 2 1 So Q divides in the ratio 2:1 So, q = m y 2 + n y 1 m + n = 2 × 2 + 1 × − 4 3 = 4 − 4 3 = 0 value of p = 7 3 and q = 0 Q. Find the coordinates of the midpoint of the line segment joining

(i) A(3, 0) and B(-5, 4) (ii) P(-11, -8) and Q(8, -2).

Solution:
(i) A(3,0) and B(-5, 4) x = 3 − 5 2 = − 2 2 = − 1 y = 0 + 4 2 = 2 point (-1,2) (ii) P(-11, -8) and Q(8, -2) x = − 11 + 8 2 = − 3 2 y = − 8 − 2 2 = − 10 2 = − 5

Q. If (2, p) is the midpoint of the line segment joining the points A(6, -5) and B(-2, 11), find the value of p.

Solution:
A(6, -5) and B(-2, 11) here mid point x = 6 − 2 2 = 2 y = − 5 + 11 2 = 3 so(2,p)=2,3 so p=3

Q. In what ratio does the point P(2, 5) divide the join of A(8, 2) and B (-6, 9) ?

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 nt h e nopen 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 parenthesesi n space t h i s space q u e s t i o n space w e space h a v e space t o space f i n d space m colon ng i v e nx equals 2 comma y equals 5 comma x subscript 1 equals 8 comma y subscript 1 equals 2 comma x subscript 2 equals negative 6 comma y subscript 2 equals 9t h e nopen parentheses 2 comma 5 close parentheses equals open parentheses fraction numerator 8 n plus negative 6 m over denominator m plus n end fraction comma fraction numerator 2 n plus 9 m over denominator m plus n end fraction close parenthesest h e n2 equals fraction numerator 8 n minus 6 m over denominator m plus n end fraction2 m plus 2 n equals 8 n minus 6 m6 n equals 8 mm over n equals 6 over 8 equals 3 over 4s o space t h e space r a t i o space i s space m colon n equals 3 colon 4

Q. In what ratio does the line x-y-2 = 0 divide the line segment joining the points A(3, -1) and B(8, 9) ?

Solution:
Let the line x − y – 2 = 0 divide the line segment joining the points A(3,-1) and B(8,9) in the ratio k:1 at P Then, by section formula the coordinates of P are x = ( m x 2 + n x 1 m + n , y = m y 2 + n y 1 m + n ) P = x = 8 k + 3 k + 1 , y = 9 k − 1 k + 1 Since, P lies on the line x − y − 2 = 0 , we have. ( 8 k + 3 k + 1 ) − ( 9 k − 1 k + 1 ) − 2 = 0 ⇒ 8 k + 3 – 9 k + 1 – 2 k − 2 = 0 ⇒ 8 k – 9 k − 2 k + 3 + 1 – 2 = 0 ⇒ − 3 k + 2 = 0 ⇒ − 3 k = − 2 ⇒ k = 2 3 So, the required ratio is 2 3 : 1 , which is equal to 2 : 3

Q. Find the lengths of the medians of a △ A B C whose vertices are A(0, -1), B(2, 1) and C(0, 3)?

Solution:
Let D be the midpoint of BC. So, the coordinates of D are D = ( 2 + 0 ) 2 , ( 1 + 3 ) 2 . ⇒ D = 2 2 , 4 2 D = ( 1 , 2 ) Let E be the midpoint of AC. So, the coordinates of E are E = ( 0 + 0 ) 2 , ( − 1 + 3 ) 2 E = ( 0 2 , 2 2 = E(0,1) Let F be the midpoint of AB. So, the coordinates of F are F = ( 0 + 2 ) 2 , ( − 1 + 1 ) 2 F = 2 2 , 0 2 = F(1,0) A D = √ ( 1 − 0 ) 2 + ( 2 − ( − 1 ) ) 2 A D = ( 1 ) 2 + ( 3 ) 2 A D = √ 1 + 9 A D = √ 1 0 u n i t s B E = √ ( 0 − 2 ) 2 + ( 1 − 1 ) 2 B E = √ ( − 2 ) 2 + ( 0 ) 2 B E = √ 4 + 0 B E = 2 u n i t s C F = √ ( 1 − 0 ) 2 + ( 0 − 3 ) 2 C F = √ ( 1 ) 2 + ( − 3 ) 2 C F = √ 1 + 9 C F = √ 10 u n i t s Therefore, the lengths of the medians: A D = √ 10 u n i t s , B E = 2 u n i t s , C F = √ 10 u n i t s Q. Find the third vertex of a △ A B C if two of its vertices are B(-3, 1) and C(0, -2), and its centroid is at the origin.
Solution:
Centroid ( X , Y ) = x 1 + x 2 + x 3 3 , y 1 + y 2 + y 3 3 ⇒ ( 0 , 0 ) = − 3 + 0 + x 3 3 , 1 − 2 + y 3 3 Solving, we get x 3 = 3 and y 3 = 1

Q. If three consecutive vertices of a parallelogram ABCD areA(1, -2), B(3, 6) and C(5, 10), find its fourth vertex D.

Solution:
Let A (1, -2), B (3, 6), C (5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D (a, b) Join AC and BD, intersecting at O We know that the diagonals of a parallelogram bisect each other Therefore, O is the midpoint of AC as well as BD Midpoint of AC = ( 1 + 5 2 , − 2 + 10 2 ) = ( 6 2 , 8 2 ) = (3, 4) Midpoint of BD = ( 3 + a 2 , 6 + b 2 ) Therefore, 3 + a 2 = 3 , 6 + b 2 = 4 ⇒ 3 + a = 6 , 6 + b = 8 ⇒ a = 6 - 3, b = 8 - 6 ⇒ a = 3 and b = 2 Therefore, the fourth vertex is D (3,2)

Q. In what ratio does y-axis divide the line segment joining the points (-4, 7) and (3, -7)?

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 w e space t a k e space t h e space o p i n t space o n space t h e space y space a x i s space i s space open parentheses 0 comma k close parentheses t h e n space w e space h a v e space t o space f i n d space m colon n g i v e n x equals 0 comma y equals k comma x subscript 1 equals negative 4 comma x subscript 2 equals 3 comma y subscript 1 equals 7 comma y subscript 2 equals negative 7 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 open parentheses 0 comma k close parentheses equals open parentheses fraction numerator negative 4 n plus 3 m over denominator m plus n end fraction comma fraction numerator 7 n plus negative 7 m over denominator m plus n end fraction close parentheses t h e n 0 equals fraction numerator negative 4 n plus 3 m over denominator m plus n end fraction minus 4 n plus 3 m equals 0 m over n equals 4 over 3 t h e space t h e space r a t i o space i s space 4 colon 3
Q. ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). If P, Q, R and S be the midpoints of AB, BC, CD and DA respectively, show that PQRS is a rhombus.

​Solution:

Here, the points P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Then The points are A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1) Co-ordinates of P = ( − 1 − 1 2 , − 1 + 4 2 ) = ( − 1 , 3 2 ) Co-ordinates of Q = ( − 1 + 5 2 , 4 + 4 2 ) = ( 2 , 4 ) Co-ordinates of R = ( 5 + 5 2 , 4 − 1 2 ) = ( 5 , 3 2 ) Co-ordinates of S = ( − 1 + 5 2 , − 1 − 1 2 ) = ( 2 , − 1 ) Now, P Q = √ ( 2 + 1 ) 2 + ( 4 − 3 2 ) 2 = √ 9 + 25 4 = √ 61 2 Q R = √ ( 5 − 2 ) 2 + ( 3 2 − 4 ) 2 = √ 9 + 25 4 = √ 61 2 R S = √ ( 5 − 2 ) 2 + ( 3 2 + 1 ) 2 = √ 9 + 25 4 = √ 61 2 S P = √ ( 2 + 1 ) 2 + ( − 1 − 3 2 ) 2 = √ 9 + 25 4 = √ 61 2 P R = √ ( 5 + 1 ) 2 + ( 3 2 − 3 2 ) 2 = √ 36 = 6 Q S = √ ( 2 − 2 ) 2 + ( − 1 − 4 ) 2 = √ 25 = 5 Thus P Q = Q R = R S = S P and P R ≠ Q S . Therefore PQRS is a rhombus

Q. Find all possible values of x for which the distance between the points

A (x, -1) and B (5, 3) is 5 units.

Solution:
We know that the distance between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) , d = √ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Given d = 5 units and points are A (x, -1) and B (5, 3) ∴ √ ( 5 − x ) 2 + ( 3 − − 1 ) 2 = 5 ( 5 − x ) 2 + ( 3 + 1 ) 2 = 25 ( 5 − x ) 2 + 16 = 25 ( 5 − x ) 2 = 25 − 16 ( 5 − x ) 2 = 9 ( 5 − x ) = √ 9 ( 5 − x ) = ± 3 x = 5 ± 3 x = 5 + 3 o r 5 − 3 x = 8 o r 2

Q. 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 √ x − 8 ) 2 + ( 2 + 2 ) 2 = √ ( x − 2 ) 2 + ( 2 + 2 ) 2 √ x 2 + 64 − 16 × x + 16 = √ x 2 + 4 − 4 × x + 16 √ x 2 − 16 × x + 80 = √ x 2 − 4 × 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= √ ( x − 8 ) 2 + ( 2 + 2 ) 2 √ ( 5 − 8 ) 2 + ( 4 ) 2 = √ 9 + 16 = √ 25 = \pm5 \) hence AB=5 Q. Show that the following points are the vertices of a square:

(i) A(3, 2), B (0, 5), C(-3, 2) and D(0, -1)

(ii) A(6, 2), B(2, 1), C(1, 5) and D(5, 6)

(iii) A(0, -2), B(3, 1), C(0, 4) and D(-3, 1)

Solution:
(i)The given points are A (3,2), B(0,5) and C(-3,2), D(0,-1). Then AB = √ ( 0 − 3 ) 2 + ( 5 − 2 ) 2 = √ ( − 3 ) 2 + ( 3 ) 2 = √ 9 + 9 = √ 18 = 3 √ 2 units BC = √ ( − 3 − 0 ) 2 + ( 2 − 5 ) 2 = √ ( − 3 ) 2 + ( − 3 ) 2 = √ 9 + 9 = √ 1 8 = 3 √ 2 units CD = √ ( 0 + 3 ) 2 + ( − 1 − 2 ) 2 = √ ( 3 ) 2 + ( − 3 ) 2 = √ 9 + 9 = √ 18 = 3 √ 2 units DA = √ ( 0 − 3 ) 2 + ( − 1 − 2 ) 2 = √ ( − 3 ) 2 + ( − 3 ) 2 = √ 9 + 9 = √ 18 = 3 √ 2 units Therefore AB = BC = CD = DA = 3 √ 2 units Also, AC = √ ( − 3 − 3 ) 2 + ( 2 − 2 ) 2 = √ ( − 6 ) 2 + ( 0 ) 2 = √ 36 = 6 units BD = √ ( 0 − 0 ) 2 + ( − 1 − 5 ) 2 = √ ( 0 ) 2 + ( − 6 ) 2 = √ 36 = 6 units Thus, diagonal AC = diagonal BD Therefore, the given points form a square. (ii) A(6, 2), B(2, 1), C(1, 5) and D(5, 6) Solution The given points are A (6,2), B(2,1) and C(1,5), D(5,6). Then A B = √ ( 2 − 6 ) 2 + ( 1 − 2 ) 2 = √ ( − 4 ) 2 + ( − 1 ) 2 = √ 16 + 1 = √ 17 units B C = √ ( 1 − 2 ) 2 + ( 5 − 1 ) 2 = √ ( − 1 ) 2 + ( − 4 ) 2 = √ 1 + 16 = √ 17 units C D = √ ( 5 − 1 ) 2 + ( 6 − 5 ) 2 = √ ( 4 ) 2 + ( 1 ) 2 = √ 16 + 1 = √ 17 units D A = √ ( 5 − 6 ) 2 + ( 6 − 2 ) 2 = √ ( 1 ) 2 + ( 4 ) 2 = √ 1 + 16 = √ 17 units Therefore A B = B C = C D = D A = √ 17 units Also, A C = √ ( 1 − 6 ) 2 + ( 5 − 2 ) 2 = √ ( − 5 ) 2 + ( 3 ) 2 = √ 25 + 9 = √ 34 units B D = √ ( 5 − 2 ) 2 + ( 6 − 1 ) 2 = √ ( 3 ) 2 + ( 5 ) 2 = √ 9 + 25 = √ 34 units Thus, diagonal AC = diagonal BD Therefore, the given points form a square. (iii)A(0, -2), B ( 3, 1), C (0, 4) and D (-3, 1) Solution The given points are P (0,-2), Q(3,1) and R(0,4), S(-3,1). Then P Q = √ ( 3 − 0 ) 2 + ( 1 + 2 ) 2 = √ ( 3 ) 2 + ( 3 ) 2 = √ 9 + 9 = √ 18 = 3 √ 2 units Q R = √ ( 0 − 3 ) 2 + ( 4 − 1 ) 2 = √ ( − 3 ) 2 + ( 3 ) 2 = √ 9 + 9 = √ 18 = 3 √ 2 units R S = √ ( − 3 − 0 ) 2 + ( 1 − 4 ) 2 = √ ( − 3 ) 2 + ( − 3 ) 2 = √ 9 + 9 = √ 18 = 3 √ 2 units S P = √ ( − 3 − 0 ) 2 + ( 1 + 2 ) 2 = √ ( − 3 ) 2 + ( 3 ) 2 = = √ 9 + 9 = √ 18 = 3 √ 2 units Therefore PQ = QR = RS = SP = 3 √ 2 units Also, P R = √ ( 0 − 0 ) 2 + ( 4 + 2 ) 2 = √ ( 0 ) 2 + ( 6 ) 2 = √ 36 = 6 units Q S = √ ( − 3 − 3 ) 2 + ( 1 − 1 ) 2 = √ ( − 6 ) 2 + ( 0 ) 2 = √ 36 = 6 units Thus, diagonal PR = diagonal QS Therefore, the given points form a square.

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2

  • Detailed Explanations: Each solution is explained step-by-step making it easier for students to understand the process of solving Coordinate Geometry problems.
  • Concept Reinforcement: The exercise focuses on applying important concepts such as the distance formula, midpoint theorem, and section formula, helping to reinforce these key ideas.
  • Error Correction: By following the detailed solutions students can learn from their mistakes and understand where they went wrong leading to better accuracy in their work.
  • Improved Accuracy: With clear explanations and methods, students can improve their accuracy in solving Coordinate Geometry problems and avoid common mistakes.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.2 FAQs

What topics does Exercise 16.2 cover?

Exercise 16.2 focuses on key Coordinate Geometry concepts such as the distance formula, midpoint theorem, and section formula. It involves solving problems related to finding distances between points, calculating midpoints, and determining coordinates of dividing points.

How can I use the solutions to improve my understanding?

By studying the solutions, you can learn the correct methods for solving problems, understand common mistakes, and see how to apply formulas effectively. Practicing with these solutions will strengthen your grasp of Coordinate Geometry.

Are the solutions aligned with the exam syllabus?

Yes, the solutions are aligned with the Class 10 Maths syllabus, ensuring that the problems and methods covered are relevant to what students will encounter in their exams.

What is Coordinate Geometry?

Coordinate Geometry, also known as Analytical Geometry, is a branch of mathematics that uses algebraic principles to solve geometric problems. It involves studying shapes and figures using a coordinate system, usually the Cartesian plane.
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