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

**Question 1. What does the equation (x – a)**^{2} + (y – b)^{2} = r^{2} become when the axes are transferred to parallel axes through the point (a–c, b)?

^{2}+ (y – b)

^{2}= r

^{2}become when the axes are transferred to parallel axes through the point (a–c, b)?

**Solution:**

We are given,

(x – a)

^{2}+ (y – b)^{2}= r^{2}Putting x = X + a – c and y = Y + b, we get,

=> ((X + a – c) – a)

^{2}+ ((Y + b ) – b)^{2}= r^{2}=> (X – c)

^{2}+ Y^{2}= r^{2}=> X

^{2}+ c^{2}– 2cX + Y^{2}= r^{2}=> X

^{2}+ Y^{2}– 2cX = r^{2}– c^{2}

Therefore the required equation is X^{2}+ Y^{2}–2cX = r^{2}– c^{2}.

**Question 2. What does the equation (a – b) (x**^{2} + y^{2}) – 2abx = 0 become if the origin is shifted to the point (ab/(a–b), 0) without rotation?

^{2}+ y

^{2}) – 2abx = 0 become if the origin is shifted to the point (ab/(a–b), 0) without rotation?

**Solution:**

We are given,

(a – b) (x

^{2}+ y^{2}) – 2abx = 0Putting x = X + [ab/(a–b)] and y = Y, we get,

=>

=>

=>

=>

=> X

^{2 }(a–b)^{2 }+ (ab)^{2 }+ 2abX (a–b)+Y^{2 }(a–b)^{2 }= 2abX (a–b)+2(ab)^{2}=> (a – b)

^{2}(X^{2}+ Y^{2}) = a^{2}b^{2}

Therefore the required equation is (a – b)^{2}(X^{2}+ Y^{2}) = a^{2}b^{2}.

**Question 3. Find what the following equations become when the origin is shifted to the point (1, 1)?**

**(i) x**^{2} + xy – 3x – y + 2 = 0

^{2}+ xy – 3x – y + 2 = 0

**Solution:**

We are given,

x

^{2}+ xy – 3x – y + 2 = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)

^{2}+ (X + 1) (Y + 1) – 3(X + 1) – (Y + 1) + 2 = 0=> X

^{2}+ 1 + 2X + XY + X + Y + 1 – 3X – 3 – Y – 1 + 2 = 0

Therefore the required equation is X^{2}+ XY = 0.

**(ii) x**^{2} – y^{2} – 2x + 2y = 0

^{2}– y

^{2}– 2x + 2y = 0

**Solution:**

We are given,

x

^{2}– y^{2}– 2x + 2y = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)

^{2}– (Y + 1)^{2}– 2(X + 1) + 2(Y + 1) = 0=> X

^{2}+ 1 + 2X – Y^{2}– 1 – 2Y – 2X – 2 + 2Y + 2 = 0=> X

^{2}– Y^{2}= 0

Therefore the required equation is X^{2}– Y^{2}= 0.

**(iii) xy – x – y + 1 = 0**

**Solution:**

We are given,

xy – x – y + 1 = 0

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

=> (X + 1) (Y + 1) – (X + 1) – (Y + 1) + 1 = 0

=> XY + X + Y + 1 – X – 1 – Y – 1 + 1 = 0

=> XY = 0

Therefore the required equation is XY = 0.

**(iv) xy – y**^{2} – x + y = 0

^{2}– x + y = 0

**Solution:**

We are given,

xy – y

^{2}– x + y = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) – (Y + 1)

^{2}– (X + 1) + (Y + 1) = 0=> XY + X + Y + 1 – Y

^{2}– 1 – 2Y – X – 1 + Y + 1 = 0=> XY – Y

^{2}= 0

Therefore, the required equation is XY – Y^{2}= 0.

**Question 4. At what point the origin be shifted so that the equation x**^{2} + xy – 3x – y + 2 = 0 does not contain any first degree term and constant term?

^{2}+ xy – 3x – y + 2 = 0 does not contain any first degree term and constant term?

**Solution:**

We are given,

x

^{2}+ xy – 3x – y + 2 = 0Suppose (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) – 3(X + a) – (Y + b) + 2 = 0=> X

^{2}+ a^{2}+ 2aX + XY + aY + bX + ab – 3X – 3a – Y – b + 2 = 0=> X

^{2}+ XY + X(2a + b – 3) + Y(a – 1) + a^{2}+ ab – 3a – b + 2 = 0As our transformed equation has no first-degree term, we have,

2a + b – 3 = 0 and a – 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 (–3, –1) remains invariant under the translation of axes when the origin is shifted to the point (–1, 3).**

**Solution:**

Here, L.H.S. = A

_{1 }= Area of the triangle with vertices (2, 3), (5, 7) and (–3, –1)=

=

=

=

= 4 sq. units

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

(X

_{1}, Y_{1}) = (2–1, 3+3) = (1, 6)(X

_{2}, Y_{2}) = (5–1, 7+3) = (4, 10)(X

_{3}, Y_{3}) = (–3–1, –1+3) = (–4, 2)Now, R.H.S. = A

_{2}= Area of the triangle with vertices (1, 6), (4, 10) and (–4, 2)=

=

=

=

= 4 sq. units

Therefore, A

_{1}= A_{2}.

Hence, proved.

**Question 6. Find what the following equations become when the origin is shifted to the point (1, 1)?**

**(i) x**^{2} + xy – 3y^{2} – y + 2 = 0

^{2}+ xy – 3y

^{2}– y + 2 = 0

**Solution:**

We are given,

x

^{2}+ xy – 3y^{2}– y + 2 = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)

^{2}+ (X + 1) (Y + 1) – 3(Y + 1)^{2}– (Y + 1) + 2 = 0=> X

^{2}+ 1 + 2X + XY + X + Y + 1 – 3Y^{2}– 3 – 6Y – Y – 1 + 2 = 0=> X

^{2}– 3Y^{2}+ XY + 3X – 6Y = 0

Therefore, the required equation is X^{2}– 3Y^{2}+ XY + 3X – 6Y = 0.

**(ii)** **xy – y**^{2} – x + y = 0

^{2}– x + y = 0

**Solution:**

We are given,

xy – y

^{2}– x + y = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) – (Y + 1)

^{2}– (X + 1)+ Y + 1 = 0=> XY + X + Y + 1 – Y

^{2}– 1 – 2Y – X – 1 + Y + 1 = 0=> XY – Y

^{2}= 0

Therefore, the required equation is XY – Y^{2}= 0.

**(iii) xy – x – y + 1 = 0**

**Solution:**

We are given,

xy – x – y + 1 = 0

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

=> (X + 1) (Y + 1) – (Y + 1) – (X + 1) + 1 = 0

=> XY + X + Y + 1 – Y – 1 – X – 1 + 1 = 0

=> XY = 0

Therefore, the required equation is XY = 0.

**(iv) x**^{2} – y^{2} – 2x + 2y = 0

^{2}– y

^{2}– 2x + 2y = 0

**Solution:**

We are given,

x

^{2}– y^{2}– 2x + 2y = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)

^{2}– (Y + 1)^{2}– 2(X + 1) + 2(Y + 1) = 0=> X

^{2}+ 1 + 2X – Y^{2}– 1 – 2Y – 2X – 2 + 2Y + 2 = 0=> X

^{2}– Y^{2}= 0

Therefore, the required equation is X^{2}– Y^{2}= 0.

**Question 7. Find the point to which the origin should be shifted after the translation of axes so that the following equations will have no first degree terms.**

**(i) x**^{2} + y^{2} – 4x – 8y + 3 = 0

^{2}+ y

^{2}– 4x – 8y + 3 = 0

**Solution:**

We are given,

x

^{2}+ y^{2}– 4x – 8y + 3 = 0Suppose (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}– 4(X + a) – 8(Y + b) + 3 = 0=> X

^{2}+ a^{2}+ 2aX + Y^{2}+ b^{2}+ 2bY – 4X – 4a – 8Y – 8b + 3 = 0=> X

^{2}+ Y^{2}+ (2a – 4)X + (2b – 8)Y + (a^{2}+ b^{2}– 4a – 8b +3) = 0As our transformed equation has no first-degree term, we have,

2a – 4 = 0 and 2b – 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) x**^{2} + y^{2} – 5x + 2y – 5 = 0

^{2}+ y

^{2}– 5x + 2y – 5 = 0

**Solution:**

We are given,

x

^{2}+ y^{2}– 5x + 2y – 5 = 0Suppose (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}– 5(X + a) + 2(Y + b) – 5 = 0=> X

^{2}+ a^{2}+ 2aX + Y^{2}+ b^{2}+ 2bY – 5X – 5a + 2Y + 2b – 5 = 0=> X

^{2}+ Y^{2}+ (2a – 5)X + (2b + 2)Y + (a^{2}+ b^{2}– 5a + 2b – 5) = 0As our transformed equation has no first-degree term, we have,

2a – 5 = 0 and 2b + 2 = 0

By solving these equations we have a = 5/2 and b = –1.

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

**(iii) x**^{2} – 12x + 4 = 0

^{2}– 12x + 4 = 0

**Solution:**

We are given,

x

^{2}– 12x + 4 = 0=> (X + a)

^{2}– 12(X + a) + 4 = 0=> X

^{2}+ a^{2}+ 2aX – 12X – 12a + 4 = 0=> X

^{2}+ (2a – 12)X + (a^{2 }– 12a + 4) = 0As our transformed equation has no first-degree term, we have,

=> 2a – 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, –2) remains invariant under the translation of axes when the origin is shifted to the point (–2, 1).**

**Solution:**

Here, L.H.S. = A

_{1}= Area of the triangle with vertices (4, 6), (7, 10) and (1, –2)=

=

=

=

= 6 sq. units

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

(X

_{1}, Y_{1}) = (4–2, 6+1) = (2, 7)(X

_{2}, Y_{2}) = (7–2, 10+1) = (5, 11)(X

_{3}, Y_{3}) = (1–2, –2+1) = (–1, –1)Now, R.H.S. = A

_{2}= Area of the triangle with vertices (2, 7), (5, 11), (–1, –1)=

=

=

=

= 6 sq. units

Therefore, A

_{1}= A_{2}.

Hence, proved.

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