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) ?

Answer:

Substituting x = X + a - c , y = Y +b in the equation (x-a)^2 + ( y-b)^2 = r^2 we get

(X + a-c-a)^2 + ( Y +b-b)^2 = r^2

\Rightarrow (X -c)^2 + ( Y )^2 = r^2

\Rightarrow X^2 + c^2 - 2Xc + Y^2 = r^2

Hence the equation gets transformed to

X^2 + Y^2 - 2Xc = r^2 - c^2

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Question 2: What does the equation ( a- b) ( x^2 + y^2)- 2abx = 0   become if the origin is shifted to the point ( \frac{ab}{a-b} , 0) without rotation?

Answer:

Substituting x = X + \frac{ab}{a-b} , y = Y +0 in the equation ( a- b) ( x^2 + y^2)- 2abx = 0 we get

(a-b) \Big[  \Big(  X + \frac{ab}{a-b}  \Big)^2 +Y^2 \Big]  - 2ab \Big(  X + \frac{ab}{a-b} \Big) = 0

\Rightarrow (a-b) \Big[ X^2 + \frac{a^2b^2}{(a-b)^2} + \frac{2X ab}{a-b} + Y^2 \Big] - 2abX - \frac{2a^2b^2}{(a-b)} = 0

\Rightarrow (a-b)(X^2+Y^2) + \frac{a^2b^2}{(a-b)} + 2abX - 2abX - \frac{2a^2b^2}{(a-b)} = 0

\Rightarrow (a-b) ( X^2 + Y^2) - \frac{a^2b^2}{(a-b)} 0

\Rightarrow (a-b) ( X^2 + Y^2) = \frac{a^2b^2}{(a-b)}

Hence the transformed equation is (a-b) ( x^2 + y^2) = \frac{a^2b^2}{(a-b)}

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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

ii) x^2 - y^2 - 2x + 2y = 0

iii) xy - x - y +1 =0

iv) xy-y^2-x+y = 0

Answer:

i)      Substituting x = X + 1 , y = Y +1 in the equation x^2 + xy - 3x - y + 2 = 0  we get

(X+1)^2 + (X+1)(Y+1) - 3(X+1) - (Y+1) + 2 = 0

\Rightarrow X^2 +1 + 2X + XY + Y + X + 1 - 3X - 3 - Y - 1 +2 = 0

\Rightarrow X^2 + XY = 0

Hence the transformed equation is x^2 + xy = 0  

ii)     Substituting x = X + 1 , y = Y +1 in the equation x^2 - y^2 - 2x + 2y = 0  we get

(X+1)^2 - (Y+1)^2 - 2(X+1) + 2(Y+1) = 0

\Rightarrow X^2 + 1 + 2X - Y^2 - 1 - 2Y - 2X - 2 + 2Y + 2 = 0

\Rightarrow X^2 - Y^2 = 0

Hence the transformed equation is x^2 -y^2 = 0  

iii)     Substituting x = X + 1 , y = Y +1 in the equation xy - x - y +1 =0  we get

(X+1)(Y+1) - (X+1) - (Y+1) +1 =0

\Rightarrow XY + Y + X + 1 - X - 1 - Y - 1 + 1 = 0

\Rightarrow XY = 0

Hence the transformed equation is xy = 0  

iv)    Substituting x = X + 1 , y = Y +1 in the equation xy-y^2-x+y = 0  we get

(X+1)(Y+1)-(Y+1)^2-(X+1)+(Y+1) = 0

\Rightarrow XY + Y + X + 1 - Y^2 - 1 - 2Y - X - 1 + Y + 1 = 0

\Rightarrow XY - Y^2 = 0

Hence the transformed equation is xy - y^2=0 

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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?

Answer:

Let the origin be shifted to (h, k) .

Therefore x = X+h , y = Y + k

Substituting x = X+h,  y = Y + k   in the equation x^2 + xy -3x-y +2 = 0 we get

(X+h)^2 + (X+h) (Y + k ) -3(X+h)- (Y + k ) +2 = 0

\Rightarrow X^2 + h^2 + 2hX + XY + hY + kX+ hk - 3X - 3h - Y - k + 2 = 0

\Rightarrow X^2 + XY + X ( 2h +k - 3) + Y ( h - 1) + h^2 + hk - 3h -k + 2 = 0

For the equation to be free of 1st degree term and constant term we get

2h + k -3 = 0,  \hspace{1,0cm} h - 1 = 0

\Rightarrow h =1 and k = 3 - 2( 1) = 1

Also h = 1 and k = 1 satisfies h^2 + hk - 3h -k + 2 = 0 .

Therefore origin should be shifted to ( 1, 1) .

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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) .

Answer:

Given \triangle ABC with vertices A(2,3), B(5,7) and C(- 3 - 1)

Area of a \triangle ABC = \frac{1}{2} | x_1 ( y_2-y_3) + x_2 ( y_3 - y_1) + x_3 ( y_1 - y_2) |

= \frac{1}{2} | 2( 7+1) + 5( -1-3) -3 ( 3-7) |

= \frac{1}{2} | 16-20+12|

= 4

Now we shift the origin to ( -1, 3)

Therefore the new vertices A'(2+1, 3-3), B'( 5+1, 7-3), C'(-3+1, -1-3) or A'(3, 0), B'( 6, 4), C'(-2, -4)

Therefore

Area of a \triangle A'B'C' = \frac{1}{2} | x_1 ( y_2-y_3) + x_2 ( y_3 - y_1) + x_3 ( y_1 - y_2) |

= \frac{1}{2} | 4( 4+4) + 6( -4-0) -2 ( 0-4) |

= \frac{1}{2} | 32 - 24 + 8|

= 4

Hence the area of the triangle would remain invariant.

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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

ii) xy - y^2 - x + y = 0

iii) xy - x -y + 1 = 0

iv)  x^2 - y^2 - 2x + 2y = 0

Answer:

i)       Substitute x = X+1, y = Y + 1 in the equation x^2 - xy - 3y^2 - y + 2 = 0 we get

(X+1)^2 + ( X+1)(Y+1) - 3 ( Y+1)^2 - ( Y+1) + 2=0

\Rightarrow X^2 + 1 + 2X + XY + Y + X + 1 - 3Y^2 - 3 - 6Y - Y -1 + 2=0

\Rightarrow X^2 + XY - 3Y^2 + 3X - 6Y = 0

Hence the equation become x^2 + xy - 3y^2 + 3x - 6y = 0 when origin is shifted to (1, 1)

ii)     Substitute x = X+1, y = Y + 1 in the equation xy - y^2 - x + y = 0 we get

(X+1)(Y+1) - ( Y+1)^2 - ( X+1) + (Y+1)=0

\Rightarrow XY + Y + X + 1 - Y^2 - 1 - 2Y - X - 1 + Y + 1 = 0

\Rightarrow XY - Y^2=0

Hence the equation become xy - y^2 = 0 when origin is shifted to (1, 1)

iii)    Substitute x = X+1, y = Y + 1 in the equation xy - x -y + 1 = 0 we get

(X+1)(Y+1) - (X+1) - ( Y+1) + 1=0

\Rightarrow XY + Y + X + 1 - X - 1 - Y - 1 + 1 = 0

\Rightarrow XY=0

Hence the equation become xy  = 0 when origin is shifted to (1, 1)

iv)     Substitute x = X+1, y = Y + 1 in the equation x^2 - y^2 - 2x + 2y = 0 we get

(X+1)^2 -(Y+1)^2 - 2 ( X+1) + 2(Y+1)=0

\Rightarrow X^1 + 1 + 2X - Y^2 - 2Y - 1 - 2X - 2 + 2 Y + 2 = 0

\Rightarrow X^2 - Y^2=0

Hence the equation become x^2 - y^2  = 0 when origin is shifted to (1, 1)

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

i) y^2 + x^2 -4x-8y +3 = 0

ii) x^2 + y^2 -5x + 2y + 5 = 0

iii) x^2 - 12x + 4 = 0

Answer:

i)       Let the origin be shifted to (h, k) .

Therefore x = X+h , y = Y + k

Substituting x = X+h,  y = Y + k   in the equation y^2 + x^2 -4x-8y +3 = 0 we get

(Y+k)^2 + ( X+h)^2 - 4 ( X +h) - 8 ( Y + k) + 3 = 0

\Rightarrow Y^2 + k^2 + 2kY + X^2 + h^2 + 2hX - 4X - 4h - 8Y -8k + 3 = 0

\Rightarrow X^2 + Y^2 + X( 2h -4) + Y ( 2k-8) + ( k^2 + h^2 - 4h - 8k + 3) = 0

For this equation to be free of terms containing X and Y we must have

2h - 4 = 0     and    2k - 8 = 0

\Rightarrow h = 2      and k = 4

Hence the origin should be shifted to ( 2, 4)

ii)      Let the origin be shifted to (h, k) .

Therefore x = X+h , y = Y + k

Substituting x = X+h,  y = Y + k   in the equation x^2 + y^2 -5x + 2y + 5 = 0 we get

(X+h)^2 + ( Y+k)^2 - 5( X+h) +2(Y+k) = 0

\Rightarrow X^1 + h^2 + 2hX + Y^2 + k^2 + 2kY - 5X -5h + 2Y + 2k = 0

\Rightarrow X^2 + Y^2 + X ( 2h -5) + Y ( 2k +2) + (h^2 + k^2 - 5h + 2k-5) = 0

For this equation to be free of terms containing X and Y we must have

2h - 5 = 0     and    2k +2 = 0

\Rightarrow h = \frac{5}{2}      and k = -1

Hence the origin should be shifted to ( \frac{5}{2}, -1)

iii)     Let the origin be shifted to (h, k) .

Therefore x = X+h , y = Y + k

Substituting x = X+h,  y = Y + k   in the equation x^2 - 12x + 4 = 0 we get

(X+h)^2 -12(X+h) + 4 = 0

\Rightarrow X^1 + h^2 + 2hX -12X - 12h + 4 = 0

\Rightarrow X^2 + X(2h-12) + (h^2 - 12h + 4)  = 0

For this equation to be free of terms containing X and Y we must have

2h - 12 = 0     \Rightarrow h = 6

Hence the origin should be shifted to ( 6, k) , k \in R

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Question 8: Verify that the area of the triangle with vertices (4,5), (7,10) and (1, -2) remains invariant under the translation of axes when the origin is shifted to the point (- 2,1) .

Answer:

Given \triangle ABC with vertices A(4,6), B(7, 10) and C(1, -2)

Area of a \triangle ABC = \frac{1}{2} | x_1 ( y_2-y_3) + x_2 ( y_3 - y_1) + x_3 ( y_1 - y_2) |

= \frac{1}{2} | 4(10+2) + 7 ( -2-6) + 1( 6-10) |

= \frac{1}{2} | 48-56-4|

= 6

Now we shift the origin to (-2, 1)

Therefore the new vertices A'(4+2, 6-1), B'( 7+2, 10-1), C'(1+2, -2-1) or A'(5, 6), B'( 9, 9), C'(3, -3)

Therefore

Area of a \triangle A'B'C' = \frac{1}{2} | x_1 ( y_2-y_3) + x_2 ( y_3 - y_1) + x_3 ( y_1 - y_2) |

= \frac{1}{2} | 6( 9+3) + 9 ( -3 -5) + 3( 5-9) |

= \frac{1}{2} | 72 - 72 -12|

= 6

Hence the area of the triangle would remain invariant.