RD Sharma Solutions for Class 11 Maths Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates

RD Sharma Solutions Class 11 Maths Chapter 22 – Download Free PDF Updated for 2023-24

RD Sharma Solutions for Class 11 Maths Chapter 22 – Brief Review of Cartesian System of Rectangular Coordinates are provided here for students to obtain in-depth knowledge of concepts and score good marks in the board exams. The RD Sharma solutions, which contain a huge number of solved examples and illustrations, provide stepwise explanations of various difficult concepts and comprise a wide variety of questions for practice. The PDF of RD Sharma Class 11 Maths Solutions is provided here. To excel in final exams, students can refer to and download this chapter PDF from the links given below.

Chapter 22 – Brief Review of Cartesian System of Rectangular Coordinates contains three exercises, and the RD Sharma Solutions provide 100% accurate answers to the questions present in each exercise. This chapter primarily deals with problems using section formulas and finding the area of a triangle. RD Sharma Solutions are completely based on the exam-oriented approach to help students acquire good marks in their board examination. Now, let us have a look at the concepts discussed in this chapter.

  • Cartesian coordinate system.
  • Distance between two points.
  • Area of a triangle.
  • Section formula.
  • Certroid, in-centre and ex-centres of a triangle.
  • Locus and equation to a locus.
  • Shifting of origin.

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

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Access answers to RD Sharma Solutions for Class 11 Maths Chapter 22 – Brief Review of Cartesian System of Rectangular Coordinates

EXERCISE 22.1 PAGE NO: 22.12

1. If the line segment joining the points P(x1, y1) and Q(x2, y2) subtends an angle α at the origin O, prove that : OP. OQ cos α = x1 x2 + y1 y2.

Solution:

Given,

Two points, P and Q, subtend an angle α at the origin, as shown in the figure:

RD Sharma Solutions for Class 11 Maths Chapter 22- image 1

From the figure, we can see that points O, P and Q form a triangle.

Clearly, in ΔOPQ we have:

RD Sharma Solutions for Class 11 Maths Chapter 22- image 2

2. The vertices of a triangle ABC are A(0, 0), B (2, -1) and C (9, 0). Find cos B.

Solution:

Given:

The coordinates of the triangle.

From the figure,

RD Sharma Solutions for Class 11 Maths Chapter 22- image 3

By using the cosine formula,

In ΔABC, we have:

RD Sharma Solutions for Class 11 Maths Chapter 22- image 4

3. Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that RD Sharma Solutions for Class 11 Maths Chapter 22- image 5, find x.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 22- image 6

RD Sharma Solutions for Class 11 Maths Chapter 22- image 7

RD Sharma Solutions for Class 11 Maths Chapter 22- image 8

24.5 = 28x – 14

28x = 38.5

x = 38.5/28

= 1.375

4. The points A (2, 0), B (9, 1), C (11, 6) and D (4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

Solution:

Given:

The coordinates of 4 points that form a quadrilateral are shown in the below figure

RD Sharma Solutions for Class 11 Maths Chapter 22- image 9

Now by using the distance formula, we have:

RD Sharma Solutions for Class 11 Maths Chapter 22- image 10

It is clear that AB ≠ BC [quad ABCD does not have all 4 sides equal.]

∴ ABCD is not a Rhombus

EXERCISE 22.2 PAGE NO: 22.18

1. Find the locus of a point equidistant from the point (2, 4) and the y-axis.

Solution:

Let P (h, k) be any point on the locus and let A (2, 4) and B (0, k).

Then, PA = PB

PA2 = PB2

RD Sharma Solutions for Class 11 Maths Chapter 22- image 11

2. Find the equation of the locus of a point which moves such that the ratio of its distance from (2, 0) and (1, 3) is 5: 4.

Solution:

Let P (h, k) be any point on the locus and let A (2, 0) and B (1, 3).

So then, PA/ BP = 5/4

PA2 = BP2 = 25/16

RD Sharma Solutions for Class 11 Maths Chapter 22- image 12

RD Sharma Solutions for Class 11 Maths Chapter 22- image 13

3. A point moves as so that the difference of its distances from (ae, 0) and (-ae, 0) is 2a, prove that the equation to its locus is

RD Sharma Solutions for Class 11 Maths Chapter 22- image 14, where b2 = a2 (e2 – 1).

Solution:

Let P (h, k) be any point on the locus and let A (ae, 0) and B (-ae, 0).

Where PA – PB = 2a

RD Sharma Solutions for Class 11 Maths Chapter 22- image 15

Now again, let us square on both sides and we get,

(eh + a)2 = (h + ae)2 + (k – 0)2

e2h2 + a2 + 2aeh = h2 + a2e2 + 2aeh + k2

h2 (e2 – 1) – k2 = a2 (e2 – 1)

RD Sharma Solutions for Class 11 Maths Chapter 22- image 16

Now let us replace (h, k) with (x, y)

The locus of a point such that the difference of its distances from (ae, 0) and (-ae, 0) is 2a.

RD Sharma Solutions for Class 11 Maths Chapter 22- image 17
Where b2 = a2 (e2 – 1)

Hence proved.

4. Find the locus of a point such that the sum of its distances from (0, 2) and (0, -2) is 6.

Solution:

Let P (h, k) be any point on the locus and let A (0, 2) and B (0, -2).

Where PA – PB = 6

RD Sharma Solutions for Class 11 Maths Chapter 22- image 18

RD Sharma Solutions for Class 11 Maths Chapter 22- image 19

5. Find the locus of a point which is equidistant from (1, 3) and x-axis.

Solution:

Let P (h, k) be any point on the locus and let A (1, 3) and B (h, 0).

Where, PA = PB

RD Sharma Solutions for Class 11 Maths Chapter 22- image 20

6. Find the locus of a point which moves such that its distance from the origin is three times is distance from x-axis.

Solution:

Let P (h, k) be any point on the locus and let A (0, 0) and B (h, 0).

Where, PA = 3PB

RD Sharma Solutions for Class 11 Maths Chapter 22- image 21

Now by squaring on both sides we get,

h2 + k2 = 9k2

h2 = 8k2

By replacing (h, k) with (x, y)

∴ The locus of the point is x2 = 8y2

EXERCISE 22.3 PAGE NO: 22.21

1. What does the equation (x – a) 2 + (y – b) 2 = r2 become when the axes are transferred to parallel axes through the point (a-c, b)?

Solution:

Given:

The equation, (x – a) 2 + (y – b) 2 = r2

The given equation (x – a)2 + (y – b)2 = r2 can be transformed into the new equation by changing x by x – a + c and y by y – b, i.e. substitution of x by x + a and y by y + b.

((x + a – c) – a)2 + ((y – b ) – b)2 = r2

(x – c)2 + y2 = r2

x2 + c2 – 2cx + y2 = r2

x2 + y2 -2cx = r2 – c2

Hence, the transformed equation is x2 + y2 -2cx = r2 – c2

2. What does the equation (a – b) (x2 + y2) – 2abx = 0 become if the origin is shifted to the point (ab / (a-b), 0) without rotation?

Solution:

Given:

The equation (a – b) (x2 + y2) – 2abx = 0

The given equation (a – b) (x2 + y2) – 2abx = 0 can be transformed into new equation by changing x by [X + ab / (a-b)] and y by Y

RD Sharma Solutions for Class 11 Maths Chapter 22- image 22

3. Find what the following equations become when the origin is shifted to the point (1, 1)?
(i) x2 + xy – 3x – y + 2 = 0
(ii) x2 – y2 – 2x + 2y = 0
(iii) xy – x – y + 1 = 0
(iv) xy – y2 – x + y = 0

Solution:

(i) x2 + xy – 3x – y + 2 = 0

Firstly let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1)2 + (x + 1) (y + 1) – 3(x + 1) – (y + 1) + 2 = 0

x2 + 1 + 2x + xy + x + y + 1 – 3x – 3 – y – 1 + 2 = 0

Upon simplification, we get,

x2 + xy = 0

∴ The transformed equation is x2 + xy = 0.

(ii) x2 – y2 – 2x + 2y = 0

Let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1)2 – (y + 1)2 – 2(x + 1) + 2(y + 1) = 0

x2 + 1 + 2x – y2 – 1 – 2y – 2x – 2 + 2y + 2 = 0

Upon simplification, we get,

x2 – y2 = 0

∴ The transformed equation is x2 – y2 = 0.

(iii) xy – x – y + 1 = 0

Let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1) (y + 1) – (x + 1) – (y + 1) + 1 = 0

xy + x + y + 1 – x – 1 – y – 1 + 1 = 0

Upon simplification, we get,

xy = 0

∴ The transformed equation is xy = 0.

(iv) xy – y2 – x + y = 0

Let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1) (y + 1) – (y + 1)2 – (x + 1) + (y + 1) = 0

xy + x + y + 1 – y2 – 1 – 2y – x – 1 + y + 1 = 0

Upon simplification, we get,

xy – y2 = 0

∴ The transformed equation is xy – y2 = 0.

4. At what point the origin be shifted so that the equation x2 + xy – 3x + 2 = 0 does not contain any first-degree term and constant term?

Solution:

Given:

The equation x2 + xy – 3x + 2 = 0

We know that the origin has been shifted from (0, 0) to (p, q)

So any arbitrary point (x, y) will also be converted as (x + p, y + q).

The new equation is:

(x + p)2 + (x + p)(y + q) – 3(x + p) + 2 = 0

Upon simplification,

x2 + p2 + 2px + xy + py + qx + pq – 3x – 3p + 2 = 0

x2 + xy + x(2p + q – 3) + y(q – 1) + p2 + pq – 3p – q + 2 = 0

For no first degree term, we have 2p + q – 3 = 0 and p – 1 = 0, and

For no constant term, we have p2 + pq – 3p – q + 2 = 0.

By solving these simultaneous equations, we have p = 1 and q = 1 from first equation.

The values p = 1 and q = 1 satisfies p2 + pq – 3p – q + 2 = 0.

Hence, the point to which the origin must be shifted is (p, q) = (1, 1).

5. Verify that the area of the triangle with vertices (2, 3), (5, 7) and (-3 -1) remains invariant under the translation of axes when the origin is shifted to the point (-1, 3).

Solution:

Given:

The points (2, 3), (5, 7), and (-3, -1).

The area of triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is

= ½ [x1(y2 – y3) + x2(y3 -y1) + x3(y1 – y2)]

The area of given triangle = ½ [2(7+1) + 5(-1-3) – 3(3-7)]

= ½ [16 – 20 + 12]

= ½ [8]

= 4

Origin shifted to point (-1, 3), and the new coordinates of the triangle are (3, 0), (6, 4), and (-2, -4) obtained from subtracting a point (-1, 3).

The new area of triangle = ½ [3(4-(-4)) + 6(-4-0) – 2(0-4)]

= ½ [24-24+8]

= ½ [8]

= 4

Since the area of the triangle before and after the translation after shifting of origin remains the same, i.e. 4.

∴ We can say that the area of a triangle is invariant to the shifting of origin.


Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 22 – Brief Review of Cartesian System of Rectangular Coordinates

Exercise 22.1 Solutions

Exercise 22.2 Solutions

Exercise 22.3 Solutions

Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 22

Q1

Why should one choose the RD Sharma Solutions for Class 11 Maths Chapter 22 from BYJU’S?

We at BYJU’S provide 100% accurate answers so that the students can confidently score good marks in the board exams. Students who have doubts while answering the textbook questions can refer to the chapter solutions to get a clear idea about the overall concepts. The solutions are framed by the expert tutors after conducting vast research on each concept. It mainly increases interest among students to learn new concepts without any difficulty.
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Students can learn the chapter and revise them on a regular basis to get a grip on the concepts. Professional teachers recommend students choose the correct study materials like the RD Sharma Solutions provided at BYJU’S to score good marks in the exam. The syllabus for the academic year should be understood before the exam to know the marks weightage of each concept covered in the chapter. Students should refer to the solutions while learning a new chapter to understand the method of answering questions as per the latest CBSE guidelines.
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Yes, RD Sharma Solutions for Class 11 Maths Chapter 22 offered by BYJU’S is accurate based on the latest guidelines set by the CBSE board. The concepts are solved in an elaborate manner to help students understand them without any difficulty. The expert tutors have created the modules passionately so that students can learn new topics efficiently. Students can access the PDF format of the syllabus as well as solutions that are available on BYJU’S website with a free download option and score well in the board exams.
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Yes, it is highly important to practice RD Sharma Solutions for Class 11 Maths Chapter 22 on a daily basis without fail to achieve high scores in exams. The solutions are explained in a detailed manner by subject matter experts to provide students with the best study source for exam preparation. Diligent practice of textbook problems following these solutions enables students not only to clear their confusion quickly but also to enhance their skills essential for effective learning from an exam perspective.
Q5

Write down the main concepts covered in RD Sharma Solutions for Class 11 Maths Chapter 22.

The main concepts covered in RD Sharma Solutions for Class 11 Maths Chapter 22 are listed below:

  • Cartesian coordinate system.
  • Distance between two points.
  • Area of a triangle.
  • Section formula.
  • Certroid, in-centre and ex-centres of a triangle.
  • Locus and equation to a locus.
  • Shifting of origin.

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