RD Sharma Solutions for Class 11 Chapter 22 – Cartesian System of Rectangular Coordinates

The RD Sharma Solutions for Class 11 Chapter 22 on the Cartesian System of Rectangular Coordinates is an essential resource for students learning coordinate geometry. This chapter is crucial for understanding higher-level mathematics and is a must-study for competitive exam preparation. It offers clear explanations, step-by-step solutions, and a variety of practice questions to help students grasp coordinate geometry concepts effectively.

Why the Cartesian System of Rectangular Coordinates is Important

The Cartesian coordinate system is foundational to mathematics, forming the basis of analytical geometry and its applications in fields like science, engineering, and technology. This chapter helps students understand how to work with coordinates, which is key to solving problems involving points, lines, and shapes on a plane. Mastering these concepts is critical for understanding more advanced topics like straight lines, circles, and conic sections. Additionally, the skills learned here—such as calculating distances, finding areas, and using section formulas—are essential for success in exams and practical applications in areas like mapping, computer graphics, and physics.

Do Check: RD Sharma Solutions for Class 6 to 12

The RD Sharma Solutions for Class 11 Chapter 22 provide detailed answers for every textbook problem. These solutions are written in simple language, making it easier for students to follow the process. The chapter also includes helpful diagrams, examples, and tips to enhance exam preparation.

By studying these RD Sharma Class 11 solutions, students will not only improve their exam performance but also develop a solid foundation in coordinate geometry, crucial for exams like JEE and others.

Download RD Sharma Solutions for Class 11 Chapter 22 – PDF

You can download the RD Sharma Class 11 Chapter 22 PDF for access to detailed solutions, solved examples, and extra practice questions. This will help you strengthen your understanding of the Cartesian system and its applications.

Before studying Chapter 22, make sure you're comfortable with basic algebra and geometry concepts, as they are essential for understanding coordinate geometry. Knowing how to plot points and perform basic arithmetic operations will help you get the most out of this chapter.

RD Sharma Class 11 Chapter 22 – Exercises

This chapter includes a range of exercises designed to help students practice and reinforce their understanding of the Cartesian coordinate system and its applications.

Question 1: Plot the points (3, 4), (–2, 5), (6, –1), and (–3, –2) on the coordinate plane.

Solution:

To plot points on a coordinate plane:

• Point (3, 4): Move 3 units right from origin, then 4 units up
• Point (–2, 5): Move 2 units left from origin, then 5 units up
• Point (6, –1): Move 6 units right from origin, then 1 unit down
• Point (–3, –2): Move 3 units left from origin, then 2 units down

The points are plotted in quadrants I, II, IV, and III respectively.

Question 2: Find the distance between the points A(2, 3) and B(5, 7).

Solution:

Using the distance formula: d = √[(x₂-x₁)² + (y₂-y₁)²]

Given points: A(2, 3) and B(5, 7)

Where x₁ = 2, y₁ = 3, x₂ = 5, y₂ = 7

d = √[(5-2)² + (7-3)²]

d = √[3² + 4²]

d = √[9 + 16]

d = √25

d = 5 units

Answer: 5 units

Question 3: Show that the points (1, 2), (2, 3), and (3, 4) are collinear.

Solution:

Three points are collinear if the area of triangle formed by them is zero.

Using the area formula: Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|

Given points: A(1, 2), B(2, 3), C(3, 4)

Area = ½|1(3-4) + 2(4-2) + 3(2-3)|

= ½|1(-1) + 2(2) + 3(-1)|

= ½|-1 + 4 - 3|

= ½|0|

= 0

Since the area is 0, the points are collinear.

Answer: The points are collinear

Question 4: Find the area of the triangle formed by the points (0, 0), (3, 4), and (6, 0).

Solution:

Using the area formula: Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|

Given points: A(0, 0), B(3, 4), C(6, 0)

Area = ½|0(4-0) + 3(0-0) + 6(0-4)|

= ½|0 + 0 + 6(-4)|

= ½|0 + 0 - 24|

= ½|-24|

= ½(24)

= 12 square units

Answer: 12 square units

Question 5: Determine the ratio in which the point (4, 7) divides the line segment joining (2, 3) and (6, 9).

Solution:

Let the point P(4, 7) divide the line segment AB in the ratio k:1

Given: A(2, 3), B(6, 9), P(4, 7)

Using section formula: P = ((k×x₂ + 1×x₁)/(k+1), (k×y₂ + 1×y₁)/(k+1))

For x-coordinate: 4 = (k×6 + 1×2)/(k+1)

4(k+1) = 6k + 2

4k + 4 = 6k + 2

4 - 2 = 6k - 4k

2 = 2k

k = 1

Verification with y-coordinate: 7 = (1×9 + 1×3)/(1+1) = 12/2 = 6 ≠ 7

Let me recalculate: If ratio is m:n, then P = ((m×x₂ + n×x₁)/(m+n), (m×y₂ + n×y₁)/(m+n))

4 = (m×6 + n×2)/(m+n) and 7 = (m×9 + n×3)/(m+n)

From first equation: 4(m+n) = 6m + 2n → 4m + 4n = 6m + 2n → 2n = 2m → n = m

So the ratio is 1:1

Answer: 1:1

Question 6: Find the coordinates of the point which divides the line segment joining (–2, 5) and (3, –5) in the ratio 2:3.

Solution:

Using section formula for internal division: P = ((m×x₂ + n×x₁)/(m+n), (m×y₂ + n×y₁)/(m+n))

Given: A(–2, 5), B(3, –5), ratio m:n = 2:3

x-coordinate = (2×3 + 3×(-2))/(2+3) = (6 - 6)/5 = 0/5 = 0

y-coordinate = (2×(-5) + 3×5)/(2+3) = (-10 + 15)/5 = 5/5 = 1

Answer: (0, 1)

Question 7: Find the coordinates of the mid-point of the line segment joining (1, 2) and (7, 4).

Solution:

Using midpoint formula: M = ((x₁+x₂)/2, (y₁+y₂)/2)

Given points: A(1, 2) and B(7, 4)

x-coordinate of midpoint = (1+7)/2 = 8/2 = 4

y-coordinate of midpoint = (2+4)/2 = 6/2 = 3

Answer: (4, 3)

Question 8: Find the centroid of the triangle with vertices at (1, 2), (3, 4), and (5, 6).

Solution:

The centroid formula: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

Given vertices: A(1, 2), B(3, 4), C(5, 6)

x-coordinate of centroid = (1+3+5)/3 = 9/3 = 3

y-coordinate of centroid = (2+4+6)/3 = 12/3 = 4

Answer: (3, 4)

Question 9: Find the coordinates of the centroid of the triangle formed by the points (0, 0), (6, 0), and (0, 8).

Solution:

Using centroid formula: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

Given vertices: A(0, 0), B(6, 0), C(0, 8)

x-coordinate of centroid = (0+6+0)/3 = 6/3 = 2

y-coordinate of centroid = (0+0+8)/3 = 8/3

Answer: (2, 8/3)

Question 10: Find the coordinates of the point which divides the line segment joining (2, –3) and (–4, 7) externally in the ratio 3:2.

Solution:

For external division, the formula is: P = ((m×x₂ - n×x₁)/(m-n), (m×y₂ - n×y₁)/(m-n))

Given: A(2, –3), B(–4, 7), ratio m:n = 3:2

x-coordinate = (3×(-4) - 2×2)/(3-2) = (-12 - 4)/1 = -16

y-coordinate = (3×7 - 2×(-3))/(3-2) = (21 + 6)/1 = 27

Answer: (–16, 27)

Question 11: Find the area of the quadrilateral formed by the points (1, 2), (4, 3), (5, 6), and (2, 5).

Solution:

Using the shoelace formula for a quadrilateral with vertices A(1, 2), B(4, 3), C(5, 6), D(2, 5):

Area = ½|x₁(y₂-y₄) + x₂(y₃-y₁) + x₃(y₄-y₂) + x₄(y₁-y₃)|

= ½|1(3-5) + 4(6-2) + 5(5-3) + 2(2-6)|

= ½|1(-2) + 4(4) + 5(2) + 2(-4)|

= ½|-2 + 16 + 10 - 8|

= ½|16|

= 8 square units

Answer: 8 square units

Question 12: Find the coordinates of the point equidistant from the three points (2, 5), (–1, 2), and (3, –2).

Solution:

Let the required point be P(x, y)

Since P is equidistant from all three points:

PA = PB = PC

PA² = PB²:

(x-2)² + (y-5)² = (x+1)² + (y-2)²

x² - 4x + 4 + y² - 10y + 25 = x² + 2x + 1 + y² - 4y + 4

-4x - 10y + 29 = 2x - 4y + 5

-6x - 6y + 24 = 0

x + y = 4 ... (1)

PA² = PC²:

(x-2)² + (y-5)² = (x-3)² + (y+2)²

x² - 4x + 4 + y² - 10y + 25 = x² - 6x + 9 + y² + 4y + 4

-4x - 10y + 29 = -6x + 4y + 13

2x - 14y + 16 = 0

x - 7y + 8 = 0 ... (2)

From (1): x = 4 - y

Substituting in (2): (4 - y) - 7y + 8 = 0

4 - y - 7y + 8 = 0

12 - 8y = 0

y = 3/2

x = 4 - 3/2 = 5/2

Answer: (5/2, 3/2)

Question 13: Find the locus of a point which moves such that its distance from the point (3, 4) is always equal to 5 units.

Solution:

Let the moving point be P(x, y)

Given: Distance from P to (3, 4) = 5

Using distance formula:

√[(x-3)² + (y-4)²] = 5

Squaring both sides:

(x-3)² + (y-4)² = 25

This is the equation of a circle with center (3, 4) and radius 5.

Answer: (x-3)² + (y-4)² = 25

Question 14: Find the coordinates of the image of the point (2, 5) when reflected over the x-axis.

Solution:

When a point is reflected over the x-axis:

The x-coordinate remains the same

The y-coordinate changes sign

Original point: (2, 5)

Image point: (2, -5)

Answer: (2, –5)

Question 15: Find the coordinates of the image of the point (–3, 6) when reflected over the y-axis.

Solution:

When a point is reflected over the y-axis:

The y-coordinate remains the same

The x-coordinate changes sign

Original point: (–3, 6)

Image point: (3, 6)

Answer: (3, 6)

Question 16: Find the coordinates of the point obtained by shifting the origin to (2, 3) for the point (5, 4).

Solution:

When the origin is shifted from (0, 0) to (h, k), the new coordinates are:

New coordinates = (x - h, y - k)

Given: Original point (5, 4), new origin (2, 3)

New x-coordinate = 5 - 2 = 3

New y-coordinate = 4 - 3 = 1

Answer: (3, 1)

Question 17: Find the new coordinates of the point (7, –2) if the axes are rotated through an angle of 90° about the origin.

Solution:

For rotation through angle θ about origin:

x' = x cos θ - y sin θ

y' = x sin θ + y cos θ

Given: Point (7, –2), θ = 90°

cos 90° = 0, sin 90° = 1

x' = 7(0) - (-2)(1) = 0 + 2 = 2

y' = 7(1) + (-2)(0) = 7 + 0 = 7

Answer: (2, 7)

Question 18: Find the area of the parallelogram formed by the points (1, 1), (2, 4), (5, 5), and (4, 2).

Solution:

Let the vertices be A(1, 1), B(2, 4), C(5, 5), and D(4, 2)

For a parallelogram, area = |AB⃗ × AD⃗|

AB⃗ = (2-1, 4-1) = (1, 3)

AD⃗ = (4-1, 2-1) = (3, 1)

Area = |1×1 - 3×3| = |1 - 9| = |-8| = 8

Alternatively, using shoelace formula for quadrilateral:

Area = ½|x₁(y₂-y₄) + x₂(y₃-y₁) + x₃(y₄-y₂) + x₄(y₁-y₃)|

= ½|1(4-2) + 2(5-1) + 5(2-4) + 4(1-5)|

= ½|2 + 8 - 10 - 16| = ½|-16| = 8

Answer: 8 square units

Question 19: Show that the points (1, 2), (3, 4), (5, 6), and (7, 8) do not form a square.

Solution:

Let A(1, 2), B(3, 4), C(5, 6), D(7, 8)

For a square, all sides must be equal and all angles must be 90°

Side lengths:

AB = √[(3-1)² + (4-2)²] = √[4 + 4] = √8 = 2√2

BC = √[(5-3)² + (6-4)²] = √[4 + 4] = √8 = 2√2

CD = √[(7-5)² + (8-6)²] = √[4 + 4] = √8 = 2√2

DA = √[(1-7)² + (2-8)²] = √[36 + 36] = √72 = 6√2

Since DA ≠ AB, the sides are not all equal.

Also, checking if points are collinear:

Slope of AB = (4-2)/(3-1) = 2/2 = 1

Slope of BC = (6-4)/(5-3) = 2/2 = 1

Since slopes are equal, the points are collinear, not forming a quadrilateral.

Answer: The points do not form a square; they are collinear

Question 20: Find the equation of the locus of a point which moves so that its distance from the x-axis is twice its distance from the y-axis.

Solution:

Let the moving point be P(x, y)

Distance from x-axis = |y|

Distance from y-axis = |x|

Given condition: |y| = 2|x|

This gives us two cases:

Case 1: y = 2x (when x and y have same sign)

Case 2: y = -2x (when x and y have opposite signs)

The locus consists of two straight lines: y = 2x and y = -2x

Answer: y = ±2x