 # RD Sharma Solutions for Class 11 Maths Exercise 22.3 Chapter 22 - Brief review of Cartesian System of Rectangular Coordinates

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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 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 

= 4

Origin shifted to point (-1, 3), 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 

= 4

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

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