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:

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

Clearly, in ΔOPQ we have:

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,

By using the cosine formula,

In ΔABC, we have:

3. Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that , find x.

Solution:

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

Now by using the distance formula, we have:

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

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

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

, 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

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)

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.

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

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

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

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

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