# Class 11 RD Sharma Solutions – Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates- Exercise 22.3

Last Updated : 28 Apr, 2021

### Question 1. What does the equation (x â€“ a)2 + (y â€“ b)2 = r2 become when the axes are transferred to parallel axes through the point (aâ€“c, b)?

Solution:

We are given,

(x â€“ a)2 + (y â€“ b)2 = r2

Putting x =  X + a â€“ c and y = Y + b, we get,

=> ((X + a â€“ c) â€“ a)2 + ((Y + b ) â€“ b)2 = r2

=> (X â€“ c)2 + Y2 = r2

=> X2 + c2 â€“ 2cX + Y2 = r2

=> X2 + Y2 â€“ 2cX = r2 â€“ c2

Therefore the required equation is X2 + Y2 â€“ 2cX = r2 â€“ c2.

### Question 2. What does the equation (a â€“ b) (x2 + y2) â€“ 2abx = 0 become if the origin is shifted to the point (ab/(aâ€“b), 0) without rotation?

Solution:

We are given,

(a â€“ b) (x2 + y2) â€“ 2abx = 0

Putting x = X + [ab/(aâ€“b)] and y = Y, we get,

=>

=>

=>

=>

=> X2 (aâ€“b)2 + (ab)2 + 2abX (aâ€“b)+Y2 (aâ€“b)2 = 2abX (aâ€“b)+2(ab)2

=> (a â€“ b)2 (X2 + Y2) = a2b2

Therefore the required equation is (a â€“ b)2 (X2 + Y2) = a2b2.

### (i) x2 + xy â€“ 3x â€“ y + 2 = 0

Solution:

We are given,

x2 + xy â€“ 3x â€“ y + 2 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)2 + (X + 1) (Y + 1) â€“ 3(X + 1) â€“ (Y + 1) + 2 = 0

=> X2 + 1 + 2X + XY + X + Y + 1 â€“ 3X â€“ 3 â€“ Y â€“ 1 + 2 = 0

Therefore the required equation is X2 + XY = 0.

### (ii) x2 â€“ y2 â€“ 2x + 2y = 0

Solution:

We are given,

x2 â€“ y2 â€“ 2x + 2y = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)2 â€“ (Y + 1)2 â€“ 2(X + 1) + 2(Y + 1) = 0

=> X2 + 1 + 2X â€“ Y2 â€“ 1 â€“ 2Y â€“ 2X â€“ 2 + 2Y + 2 = 0

=> X2 â€“ Y2 = 0

Therefore the required equation is X2 â€“ Y2 = 0.

### (iii) xy â€“ x â€“ y + 1 = 0

Solution:

We are given,

xy â€“ x â€“ y + 1 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) â€“ (X + 1) â€“ (Y + 1) + 1 = 0

=> XY + X + Y + 1 â€“ X â€“ 1 â€“ Y â€“ 1 + 1 = 0

=> XY = 0

Therefore the required equation is XY = 0.

### (iv) xy â€“ y2 â€“ x + y = 0

Solution:

We are given,

xy â€“ y2 â€“ x + y = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) â€“ (Y + 1)2 â€“ (X + 1) + (Y + 1) = 0

=> XY + X + Y + 1 â€“ Y2 â€“ 1 â€“ 2Y â€“ X â€“ 1 + Y + 1 = 0

=> XY â€“ Y2 = 0

Therefore, the required equation is XY â€“ Y2 = 0.

### Question 4. At what point the origin be shifted so that the equation x2 + xy â€“ 3x â€“ y + 2 = 0 does not contain any first degree term and constant term?

Solution:

We are given,

x2 + xy â€“ 3x â€“ y + 2 = 0

Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)2 + (X + a)(Y + b) â€“ 3(X + a) â€“ (Y + b) + 2 = 0

=> X2 + a2 + 2aX + XY + aY + bX + ab â€“ 3X â€“ 3a â€“ Y â€“ b + 2 = 0

=> X2 + XY + X(2a + b â€“ 3) + Y(a â€“ 1) + a2 + ab â€“ 3a â€“ b + 2 = 0

As our transformed equation has no first-degree term, we have,

2a + b â€“ 3 = 0 and a â€“ 1 = 0

By solving these equations we have a = 1 and b = 1.

Therefore, the origin has been shifted to (1,1) from (0,0).

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

Solution:

Here, L.H.S. = A1 = Area of the triangle with vertices (2, 3), (5, 7) and (â€“3, â€“1)

= 4 sq. units

As the origin shifted to point (â€“1, 3), the new coordinates of the triangle are:

(X1, Y1) = (2â€“1, 3+3) = (1, 6)

(X2, Y2) = (5â€“1, 7+3) = (4, 10)

(X3, Y3) = (â€“3â€“1, â€“1+3) = (â€“4, 2)

Now, R.H.S. = A2 = Area of the triangle with vertices (1, 6), (4, 10) and (â€“4, 2)

= 4 sq. units

Therefore, A1 = A2.

Hence, proved.

### (i) x2 + xy â€“ 3y2 â€“ y + 2 = 0

Solution:

We are given,

x2 + xy â€“ 3y2 â€“ y + 2 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)2 + (X + 1) (Y + 1) â€“ 3(Y + 1)2 â€“ (Y + 1) + 2 = 0

=> X2 + 1 + 2X + XY + X + Y + 1 â€“ 3Y2 â€“ 3 â€“ 6Y â€“ Y â€“ 1 + 2 = 0

=> X2 â€“ 3Y2 + XY + 3X â€“ 6Y = 0

Therefore, the required equation is X2 â€“ 3Y2 + XY + 3X â€“ 6Y = 0.

### (ii)xy â€“ y2 â€“ x + y = 0

Solution:

We are given,

xy â€“ y2 â€“ x + y = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) â€“ (Y + 1)2 â€“ (X + 1)+ Y + 1 = 0

=> XY + X + Y + 1 â€“ Y2 â€“ 1 â€“ 2Y â€“ X â€“ 1 + Y + 1 = 0

=> XY â€“ Y2 = 0

Therefore, the required equation is XY â€“ Y2 = 0.

### (iii) xy â€“ x â€“ y + 1 = 0

Solution:

We are given,

xy â€“ x â€“ y + 1 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) â€“ (Y + 1) â€“ (X + 1) + 1 = 0

=> XY + X + Y + 1 â€“ Y â€“ 1 â€“ X â€“ 1 + 1 = 0

=> XY = 0

Therefore, the required equation is XY = 0.

### (iv) x2 â€“ y2 â€“ 2x + 2y = 0

Solution:

We are given,

x2 â€“ y2 â€“ 2x + 2y = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)2 â€“ (Y + 1)2 â€“ 2(X + 1) + 2(Y + 1) = 0

=> X2 + 1 + 2X â€“ Y2 â€“ 1 â€“ 2Y â€“ 2X â€“ 2 + 2Y + 2 = 0

=> X2 â€“ Y2 = 0

Therefore, the required equation is X2 â€“ Y2 = 0.

### (i) x2 + y2 â€“ 4x â€“ 8y + 3 = 0

Solution:

We are given,

x2 + y2 â€“ 4x â€“ 8y + 3 = 0

Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)2 + (Y + b)2 â€“ 4(X + a) â€“ 8(Y + b) + 3 = 0

=> X2 + a2 + 2aX + Y2 + b2 + 2bY â€“ 4X â€“ 4a â€“ 8Y â€“ 8b + 3 = 0

=> X2 + Y2 + (2a â€“ 4)X + (2b â€“ 8)Y + (a2 + b2 â€“ 4a â€“ 8b +3) = 0

As our transformed equation has no first-degree term, we have,

2a â€“ 4 = 0 and 2b â€“ 8 = 0

By solving these equations we have a = 2 and b = 4.

Therefore, the origin has been shifted to (2,4) from (0,0).

### (ii) x2 + y2 â€“ 5x + 2y â€“ 5 = 0

Solution:

We are given,

x2 + y2 â€“ 5x + 2y â€“ 5 = 0

Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)2 + (Y + b)2 â€“ 5(X + a) + 2(Y + b) â€“ 5 = 0

=> X2 + a2 + 2aX + Y2 + b2 + 2bY â€“ 5X â€“ 5a + 2Y + 2b â€“ 5 = 0

=> X2 + Y2 + (2a â€“ 5)X + (2b + 2)Y + (a2 + b2 â€“ 5a + 2b â€“ 5) = 0

As our transformed equation has no first-degree term, we have,

2a â€“ 5 = 0 and 2b + 2 = 0

By solving these equations we have a = 5/2 and b = â€“1.

Therefore, the origin has been shifted to (5/2, â€“1) from (0,0).

### (iii) x2 â€“ 12x + 4 = 0

Solution:

We are given,

x2 â€“ 12x + 4 = 0

Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)2 â€“ 12(X + a) + 4 = 0

=> X2 + a2 + 2aX â€“ 12X â€“ 12a + 4 = 0

=> X2 + (2a â€“ 12)X + (a2 â€“ 12a + 4) = 0

As our transformed equation has no first-degree term, we have,

=> 2a â€“ 12 = 0

=> a = 6

Therefore, the origin has been shifted to (6,b) from (0,0) where b is any arbitrary value.

### Question 8. Verify that the area of the triangle with vertices (4, 6), (7, 10), and (1, â€“2) remains invariant under the translation of axes when the origin is shifted to the point (â€“2, 1).

Solution:

Here, L.H.S. = A1 = Area of the triangle with vertices (4, 6), (7, 10) and (1, â€“2)

= 6 sq. units

As the origin shifted to point (â€“2, 1), the new coordinates of the triangle are:

(X1, Y1) = (4â€“2, 6+1) = (2, 7)

(X2, Y2) = (7â€“2, 10+1) = (5, 11)

(X3, Y3) = (1â€“2, â€“2+1) = (â€“1, â€“1)

Now, R.H.S. = A2 = Area of the triangle with vertices (2, 7), (5, 11), (â€“1, â€“1)

= 6 sq. units

Therefore, A1 = A2.

Hence, proved.

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