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By rohit.pandey1
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Updated on 1 Jul 2025, 12:04 IST
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.
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.
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.
This chapter includes a range of exercises designed to help students practice and reinforce their understanding of the Cartesian coordinate system and its applications.
Solution:
To plot points on a coordinate plane:
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The points are plotted in quadrants I, II, IV, and III respectively.
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
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
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
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
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)
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)
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)
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)
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)
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
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)
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
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)
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)
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)
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)
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
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
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
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The Cartesian coordinate system is a method for representing points in a plane using two perpendicular axes, the x-axis and y-axis. Each point is identified by an ordered pair (x, y).
The distance between two points (x1, y1) and (x2, y2) is calculated using the formula:
Distance = √[(x2 - x1)² + (y2 - y1)²]
The section formula is used to find the coordinates of a point that divides a line segment joining two points (x1, y1) and (x2, y2) internally or externally in a given ratio.
Internal Division: If a point divides the line segment in the ratio m:n internally, then:
Coordinates = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))
External Division: If a point divides the line segment in the ratio m:n externally, then:
Coordinates = ((mx2 - nx1)/(m - n), (my2 - ny1)/(m - n))
The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by:
Area = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Shifting the origin to a new point (h, k) changes the coordinates of a point (x, y) to (x - h, y - k).
If the original coordinates are (x, y) and the origin is shifted to (h, k), then:
New coordinates = (x - h, y - k)
For a triangle with vertices (x1, y1), (x2, y2), and (x3, y3):
Where a, b, c are the lengths of the sides opposite to vertices A, B, C respectively.