RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1
Here, we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1. Students can view these RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 before exams for better understanding.
Ananya Gupta28 Sept, 2025
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RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1: RS Aggarwal Solutions for Class 10 Maths Chapter 16 Co-ordinate Geometry Exercise 16.1 provide detailed explanations and step-by-step solutions to help students understand the fundamentals of coordinate geometry.
This exercise focuses on the basics of plotting points on the Cartesian plane, understanding the concepts of the x-axis and y-axis, and calculating the distance between two points using the distance formula.These solutions are an excellent resource for reinforcing classroom learning and preparing for exams.
RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 Overview
RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 prepared by subject experts from Physics Wallah provide a comprehensive overview of coordinate geometry.With clear explanations and step-by-step instructions, these solutions help students grasp the basics of coordinate geometry, making it easier to solve related problems accurately and confidently.
RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 PDF
The PDF link for RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 is available below.It includes step-by-step explanations to help students understand and master the concepts, making it an invaluable resource for exam preparation and homework help.
RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 PDF
RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1
Below we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 for the ease of the students –Q. Find the distance between the points :
(i) A(9, 3) and B (15, 11) (ii) A (7, -4) and B(-5, 1)
(iii) A(-6, -4) and B(9, -12) (iv) A (1, -3 ) and B (4, -6)
(v) P (a+b, a-b) and Q (a-b, a+b)
(vi) P (a sin α, a cos α) and Q (a cos α, - a sin α)
Solution:
(i) A(5, -12) (ii) B (-5, 5) (iii) C (-4, -6).
Solution:
(i) The distance of point (5,-12) from the origin is
Origin (0,0) point (5,-12) √[5²+(-12)²] = √25+144 = √169=13(ii) origin (0,0)
point (-5,5) √(25+25)=√50=5√2(iii) origin (0,0)
point (-4,-6) √16+36=√52=2√13 Q. Find all possible values of y for which the distance between the pointsA (2, -3) and B (10, y) is 10 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 = 10 units and points are A (2, -3) and B (10, y) ∴ √ ( 10 − 2 ) 2 + ( y − − 3 ) 2 = 10 8 2 + ( y + 3 ) 2 = 100 ( y + 3 ) 2 + 64 = 100 ( y + 3 ) 2 = 100 − 64 ( y + 3 ) 2 = 36 ( y + 3 ) = √ 36 ( y + 3 ) = ± 6 y = ± 6 − 3 y = 6 − 3 o r − 6 − 3 y = 3 o r − 9
Q. Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.
Solution:
Coordinates of P and Q are (x, 4) and (9, 10) and PQ = 10 units 
→ ( P Q ) 2 = ( 10 ) 2 u n i t s = 100 u n i t s ⇒ 9 – x 2 + 10 – 4 2 = 100 ⇒ 81–x2–18\timesx+36=100 → 81 – x 2 – 18 × x + 36 = 100 → x 2 – ( 18 × x ) + 17 = 0 →(x–1)(x–17)=0 →x=1orx=17 Hence the value of x is either 1 or 17.
→ ( P Q ) 2 = ( 10 ) 2 u n i t s = 100 u n i t s ⇒ 9 – x 2 + 10 – 4 2 = 100 ⇒ 81–x2–18\timesx+36=100 → 81 – x 2 – 18 × x + 36 = 100 → x 2 – ( 18 × x ) + 17 = 0 →(x–1)(x–17)=0 →x=1orx=17 Hence the value of x is either 1 or 17.
Q. Find the coordinates of the point on x-axis which is equidistant from the points (-2, 5) and (2, -3).
Solution:
On x axis, the y coordinate is 0 So, let us assume the point to be (x,0) Distance from (-2,5) = Distance from (2,-3) ( x + 2 ) 2 + 25 = ( x − 2 ) 2 + 9 16= -8x x = -2 So, the point required is (-2,0)
Q. Find points on the x-axis, each of which is at a distance of 10 units from the point A(11, -8).
Solution:
√ ( ( x − 11 ) 2 + ( 0 + 8 ) 2 ) = 10 Square both sides and expand x 2 − 22 x + 121 + 64 = 100 x 2 − 22 x + 85 = 0 ( x − 17 ) ( x − 5 ) = 0 x = 17 o r x = 5
Q. Find the point on the y-axis which is equidistant from the points A(6, 5) and B(-4, 3).
Solution:
Let the point be (0 , y) So, √ 6 2 + ( 5 − y ) 2 = √ ( − 4 ) 2 + ( 3 − y ) 2 On squaring both sides , we get 36 + 25 + y 2 − 10 y = 16 + 9 + y 2 − 6 y 61 − 10 y = 25 − 6 y 10 y − 6 y = 61 − 25 4 y = 36 So, y = 9 So, point = (0, 9)
If P is equidistant from point A and B then,AP:PB=1:1 Distance AP = √ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = √ ( x − 5 ) 2 + ( y − 1 ) 2 A P 2 = x 2 + 25 − 10 x + y 2 + 1 − 2 y A P 2 = x 2 + y 2 + 26 − 10 x − 2 y Distance B P = √ ( − 1 − x ) 2 + ( 5 − y ) 2 B P 2 = 1 + x 2 + 2 x + 25 + y 2 − 10 y B P 2 = 26 + x 2 + y 2 + 2 x − 10 y Since P is the midpoint. A P 2 = B P 2 x 2 + y 2 + 26 − 10 x − 2 y = x 2 + y 2 + 2 x − 10 y + 26 x 2 − x 2 + y 2 − y 2 + 26 − 26 − 10 x + 2 x = − 10 y + 2 y − 12 x = − 8 y − 3 x = − 2 y 3 x = 2 y hence proved
Q. If the point P(x, y) is equidistant from the points A(5, 1) and B(-1, 5) prove that 3x = 2y.
Solution:
If P is equidistant from point A and B then,AP:PB=1:1 Distance AP = √ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = √ ( x − 5 ) 2 + ( y − 1 ) 2 A P 2 = x 2 + 25 − 10 x + y 2 + 1 − 2 y A P 2 = x 2 + y 2 + 26 − 10 x − 2 y Distance B P = √ ( − 1 − x ) 2 + ( 5 − y ) 2 B P 2 = 1 + x 2 + 2 x + 25 + y 2 − 10 y B P 2 = 26 + x 2 + y 2 + 2 x − 10 y Since P is the midpoint. A P 2 = B P 2 x 2 + y 2 + 26 − 10 x − 2 y = x 2 + y 2 + 2 x − 10 y + 26 x 2 − x 2 + y 2 − y 2 + 26 − 26 − 10 x + 2 x = − 10 y + 2 y − 12 x = − 8 y − 3 x = − 2 y 3 x = 2 y hence proved
Q. Find the coordinates of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5).
Solution:
Let the required point be P(x,y). Then A P = B P = C P That is, ( A P ) 2 = ( B P ) 2 = ( C P ) 2 This means, ( A P ) 2 = ( B P ) 2 ⇒ ( x − 5 ) 2 + ( y − 3 ) 2 = ( x − 5 ) 2 + ( y + 5 ) 2 ⇒ x 2 − 10 x + 25 + y 2 – 6 y + 9 = x 2 − 10 x + 25 + y 2 + 10 y + 25 ⇒ x 2 + y 2 − 10 x − 6 y + 34 = x 2 + y 2 – 10 x + 10 y + 50 ⇒ x 2 + y 2 − 10 x − 6 y – x 2 – y 2 + 10 x – 10 y = 50 − 34 ⇒ − 16 y = 16 ⇒ y = − 16 / 16 = − 1 And ( B P ) 2 = ( C P ) 2 ⇒ ( x − 5 ) 2 + ( y + 5 ) 2 = ( x − 1 ) 2 + ( y + 5 ) 2 ⇒ x 2 − 10 x + 25 + y 2 + 10 y + 25 = x 2 − 2 x + 1 + y 2 + 10 y + 25 ⇒ x 2 + y 2 − 10 x + 10 y + 50 = x 2 + y 2 – 2 x + 10 y + 26 ⇒ x 2 + y 2 − 10 x + 10 y – x 2 – y 2 + 2 x – 10 y = 26 − 50 ⇒ − 8 x = − 24 ⇒ y = − 24 / − 8 = 3 Hence the required point is (3,-1)
given distance AC=distance BC
then 
+ 25 - 10x + 
+ 1 - 2y = 
+ 1 + 2x + 
+ 25 - 10y - 12x = -8y 3x = 2y Hence proved Q. Using the distance formula, show that the given points are collinear:
Q. If the point C(-2, 3) is equidistant from the points A(3, -1) and B(x, 8), find the values of x. Also, find the distance BC.
Solution:
given distance AC=distance BCQ. (i) If the point P(2, 2) is equidistant from the points (a+b, b-a) and (a-b, a+b), prove that bx = ay. (ii) If the distances of P(x, y) from A(5, 1) and B (-1, 5) are equal then prove that 3x = 2y.
Solution:
(i)
Distance between the points (x, y) and (a+b, b-a) & (a-b, a+b) is equal ⇒ √ [ x − ( a + b ) ] 2 + [ y − ( b − a ) ] 2 = √ [ x − ( a − b ) ] 2 + [ y − ( a + b ) ] 2 ⇒ x 2 + ( a + b ) 2 − 2 x ( a + b ) + y 2 + ( b − a ) 2 − 2 y ( b − a ) = x 2 + ( a − b ) 2 − 2 x ( a − b ) + y 2 + ( a + b ) 2 − 2 y ( a + b ) ⇒ − 2 a x − 2 b x − 2 b y + 2 a y = − 2 a x + 2 b x − 2 a y − 2 b y ⇒ a y − b x = b x − a y ⇒ 2 a y = 2 b x ⇒ b x = a y(ii)
It is given that P is equidistant from A and B. So, PA = PB Using distance formula, then + 25 - 10x + + 1 - 2y = + 1 + 2x + + 25 - 10y - 12x = -8y 3x = 2y Hence proved Q. Using the distance formula, show that the given points are collinear:
(i) (1, -1), (5, 2) and 9, 5 (ii) (6, 9), (0, 1) and (-6, -7)
(iii) (-1, -1), (2, 3) and (8, 11) (iv) (-2, 5), (0, 1) and (2, 3).
Solution:
Q. Show that the points A(7 , 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.
Solution:
The given points are A(7,10), B(-2,5) and C(3,-4) AB = √ ( − 2 − 7 ) 2 + ( 5 − 10 ) 2 = √ ( − 9 ) 2 + ( − 5 ) 2 = √ 81 + 25 =√106 units BC = √ ( 3 − ( − 2 ) ) 2 + ( − 4 − 5 ) 2 = √ ( 5 ) 2 + ( − 9 ) 2 = √ 25 + 81 =√106 units AC = √ ( 3 − 7 ) 2 + ( − 4 − 10 ) 2 = √ ( − 4 ) 2 + ( − 14 ) 2 = √ 16 + 196 = √ 212 u n i t s Since, AB and BC are equal, they form the vertices of an isosceles triangle Also, ( A B ) 2 + ( B C ) 2 = √ ( 106 ) 2 + √ ( 106 ) 2 = √212 and ( A C ) 2 = ( √ 212 ) 2 = 212 Thus, ( A B ) 2 + ( B C ) 2 = ( A C ) 2 This shows that ΔABC is right angled at B Therefore, the given points A(7,10), B(-2,5) and C(3,-4) are the vertices of an isosceles right-angled triangle.
Q. Show that the points (-3 , 3), (3, 3) and ( − 3 √ 3 , 3 √ 3 ) are the vertices of an equilateral triangle.
Solution:
The given points are A (-3,-3), B(3,3) and C(−3√3,3√3). Now, AB =√[(−3−3)^2+(−3−3)^2] = √[(−6)^2+(−6)^2] = √(36+36)=√72 = 6√2 BC = √[(3+3√3)^2+(3−3√3)^2] =√[9+27+18√3+9+27−18√3] = √72=6√2 AC =√[(−3+3√3)^2+(−3−3√3)^2] =√[9+27−18√3+9+27+18√3] = √72= 6√2 AB=BC=AC Hence the given points are the vertices of an equilateral triangle Q. In the given figure P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Then y equals
option a
(a) 2 (b) 4 (c) -4 (d) − 5 2
Solution:
option a Q . The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in square units is (a) 20 (b) 12 (c) 6 (d) 16
Solution:
Area of the triangle = 1 2 | x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1 ) + x 3 ( y 1 − y 2 ) | = 1 2 | 5 ( 0 − 4 ) + 8 ( 4 − 0 ) + 8 ( 0 − 0 ) | = 1 2 | − 20 + 32 | = 6 s q . u n i t s Q. If A(-6, 7) and B(-1, -5) are two given points then the distance 2AB is
(a) 13 (b) 26 (c) 169 (d) 238
Solution:
Q. The distance of P(3, 4) from the x-axis is (a) 3 units (b) 4 units (c) 5 units (d) 1 unit
Solution:
A pair of Cartesian coordinates (a,b) identifies the position of a point. That point is defined as being on the line x=a where it crosses the line y=b. So by definition, that point is |b| from the X axis, and |a| from the y axis. so the answer is 4 unit option b is correct
Q. If P(-1, 1) is the midpoint of the line segment joining A(-3, b) and B(1, b+4) then b=? (a) 1 (b) -1 (c) 2 (d) 0
Solution:
The given ports are A(-3,b) and B(1, b+4) y = b + b + 4 2 1 = 2 b + 4 2 1 = b + 2 b = − 1 Option B
Q. If A(4, 2), B(6, 5) and C(1, 4) be the vertices of △ A B C and AD is a median, then the coordinates of D are (a) ( 5 2 , 3 ) (b) ( 5 , 7 2 ) (c) ( 7 2 , 9 2 ) (d) none of these
Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC. D(x,y) = {(6+1)/2 , (5+4)/2} = {7/2 ,9/2} (c) (7/2,9/2) is correct answer
Solution:
Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC. D(x,y) = {(6+1)/2 , (5+4)/2} = {7/2 ,9/2} (c) (7/2,9/2) is correct answer
Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1
- Clear Explanations: The solutions provide detailed step-by-step explanations making complex concepts in coordinate geometry easy to understand.
- Enhanced Understanding: By working through these solutions students can deepen their understanding of plotting points understanding the Cartesian plane and using the distance formula.
- Exam Preparation: These solutions align with the Class 10 syllabus helping students prepare effectively for exams by focusing on the key concepts and types of questions that may appear.
- Confidence Building: With comprehensive solutions and explanations, students can build confidence in their ability to tackle coordinate geometry problems, leading to better performance in tests and exams.
RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 FAQs
What is coordinate geometry?
Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. It involves plotting points, lines, and shapes on a Cartesian plane.
What is a Cartesian plane?
A Cartesian plane is a two-dimensional plane defined by a horizontal axis (x-axis) and a vertical axis (y-axis). The point where these axes intersect is called the origin, denoted by (0, 0).
How do you plot a point on the Cartesian plane?
To plot a point, you need an ordered pair of numbers (x, y). The x-coordinate represents the horizontal distance from the origin, and the y-coordinate represents the vertical distance. For example, to plot the point (3, 4), move 3 units to the right along the x-axis and 4 units up along the y-axis.
How do you determine if two lines are perpendicular?
Two lines are perpendicular if the product of their slopes is -1.
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