\displaystyle \textbf{Question 1: }\text{What does the equation }(x-a)^2+(y-b)^2=r^2\text{ become when the axes are}
\displaystyle \text{transferred to parallel axes through the point }(a-c,b)?
\displaystyle \text{Answer:}
\displaystyle \text{The new origin is }(a-c,b).
\displaystyle \therefore x=X+a-c,\qquad y=Y+b.
\displaystyle \text{Substituting these in }(x-a)^2+(y-b)^2=r^2,\text{ we get}
\displaystyle (X+a-c-a)^2+(Y+b-b)^2=r^2
\displaystyle \Rightarrow (X-c)^2+Y^2=r^2
\displaystyle \Rightarrow X^2-2cX+c^2+Y^2=r^2
\displaystyle \Rightarrow X^2+Y^2-2cX=r^2-c^2
\displaystyle \therefore \text{The transformed equation is }X^2+Y^2-2cX=r^2-c^2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What does the equation }(a-b)(x^2+y^2)-2abx=0\text{ become if the origin}
\displaystyle \text{is shifted to the point }\left(\frac{ab}{a-b},0\right)\text{ without rotation?}
\displaystyle \text{Answer:}
\displaystyle \text{The new origin is }\left(\frac{ab}{a-b},0\right).
\displaystyle \therefore x=X+\frac{ab}{a-b},\qquad y=Y.
\displaystyle \text{Substituting these in }(a-b)(x^2+y^2)-2abx=0,\text{ we get}
\displaystyle (a-b)\left[\left(X+\frac{ab}{a-b}\right)^2+Y^2\right]
\displaystyle \qquad-2ab\left(X+\frac{ab}{a-b}\right)=0
\displaystyle \Rightarrow (a-b)\left[X^2+\frac{2abX}{a-b}+\frac{a^2b^2}{(a-b)^2}+Y^2\right]
\displaystyle \qquad-2abX-\frac{2a^2b^2}{a-b}=0
\displaystyle \Rightarrow (a-b)(X^2+Y^2)+2abX+\frac{a^2b^2}{a-b}
\displaystyle \qquad-2abX-\frac{2a^2b^2}{a-b}=0
\displaystyle \Rightarrow (a-b)(X^2+Y^2)-\frac{a^2b^2}{a-b}=0
\displaystyle \Rightarrow (a-b)(X^2+Y^2)=\frac{a^2b^2}{a-b}
\displaystyle \Rightarrow X^2+Y^2=\frac{a^2b^2}{(a-b)^2}
\displaystyle \therefore \text{The transformed equation is }X^2+Y^2=\frac{a^2b^2}{(a-b)^2}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find what the following equations become when the origin is shifted to the point }(1,1):
\displaystyle \text{(i) }x^2+xy-3x-y+2=0
\displaystyle \text{(ii) }x^2-y^2-2x+2y=0
\displaystyle \text{(iii) }xy-x-y+1=0
\displaystyle \text{(iv) }xy-y^2-x+y=0
\displaystyle \text{Answer:}
\displaystyle \text{Since the origin is shifted to }(1,1),\text{ we substitute }x=X+1,\ y=Y+1.
\displaystyle \text{(i) Given }x^2+xy-3x-y+2=0
\displaystyle \therefore (X+1)^2+(X+1)(Y+1)-3(X+1)-(Y+1)+2=0
\displaystyle \Rightarrow X^2+2X+1+XY+X+Y+1-3X-3-Y-1+2=0
\displaystyle \Rightarrow X^2+XY=0
\displaystyle \therefore \text{The transformed equation is }X^2+XY=0.
\displaystyle \text{(ii) Given }x^2-y^2-2x+2y=0
\displaystyle \therefore (X+1)^2-(Y+1)^2-2(X+1)+2(Y+1)=0
\displaystyle \Rightarrow X^2+2X+1-Y^2-2Y-1-2X-2+2Y+2=0
\displaystyle \Rightarrow X^2-Y^2=0
\displaystyle \therefore \text{The transformed equation is }X^2-Y^2=0.
\displaystyle \text{(iii) Given }xy-x-y+1=0
\displaystyle \therefore (X+1)(Y+1)-(X+1)-(Y+1)+1=0
\displaystyle \Rightarrow XY+X+Y+1-X-1-Y-1+1=0
\displaystyle \Rightarrow XY=0
\displaystyle \therefore \text{The transformed equation is }XY=0.
\displaystyle \text{(iv) Given }xy-y^2-x+y=0
\displaystyle \therefore (X+1)(Y+1)-(Y+1)^2-(X+1)+(Y+1)=0
\displaystyle \Rightarrow XY+X+Y+1-Y^2-2Y-1-X-1+Y+1=0
\displaystyle \Rightarrow XY-Y^2=0
\displaystyle \therefore \text{The transformed equation is }XY-Y^2=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{To what point should the origin be shifted so that the equation}
\displaystyle x^2+xy-3x-y+2=0\text{ contains neither first-degree terms nor a constant term?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the origin be shifted to }(h,k).
\displaystyle \therefore x=X+h,\qquad y=Y+k.
\displaystyle \text{Substituting these in }x^2+xy-3x-y+2=0,\text{ we get}
\displaystyle (X+h)^2+(X+h)(Y+k)-3(X+h)-(Y+k)+2=0
\displaystyle \Rightarrow X^2+2hX+h^2+XY+kX+hY+hk-3X-3h-Y-k+2=0
\displaystyle \Rightarrow X^2+XY+X(2h+k-3)+Y(h-1)
\displaystyle \qquad+h^2+hk-3h-k+2=0
\displaystyle \text{For the equation to contain no first-degree terms,}
\displaystyle 2h+k-3=0\qquad\text{and}\qquad h-1=0.
\displaystyle \Rightarrow h=1
\displaystyle \therefore k=3-2h=3-2=1.
\displaystyle \text{Also, for }h=1\text{ and }k=1,
\displaystyle h^2+hk-3h-k+2=1+1-3-1+2=0.
\displaystyle \therefore \text{The origin should be shifted to }(1,1).
\displaystyle \text{The transformed equation is }X^2+XY=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Verify that the area of the triangle with vertices }(2,3),\ (5,7)\text{ and}
\displaystyle (-3,-1)\text{ remains invariant under translation of axes when the origin is shifted}
\displaystyle \text{to the point }(-1,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let the vertices of }\triangle ABC\text{ be }A(2,3),\ B(5,7)\text{ and }C(-3,-1).
\displaystyle \text{Area of }\triangle ABC
\displaystyle =\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|
\displaystyle =\frac{1}{2}\left|2(7+1)+5(-1-3)-3(3-7)\right|
\displaystyle =\frac{1}{2}|16-20+12|
\displaystyle =\frac{1}{2}\times8=4\text{ square units.}
\displaystyle \text{Now, the origin is shifted to }(-1,3).
\displaystyle \therefore x=X-1,\qquad y=Y+3,
\displaystyle \text{or }X=x+1,\qquad Y=y-3.
\displaystyle \therefore A'(2+1,3-3)=(3,0)
\displaystyle B'(5+1,7-3)=(6,4)
\displaystyle C'(-3+1,-1-3)=(-2,-4)
\displaystyle \text{Area of }\triangle A'B'C'
\displaystyle =\frac{1}{2}\left|3(4+4)+6(-4-0)-2(0-4)\right|
\displaystyle =\frac{1}{2}|24-24+8|
\displaystyle =\frac{1}{2}\times8=4\text{ square units.}
\displaystyle \therefore \text{The area of the triangle remains invariant under the translation of axes.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find what the following equations become when the origin is shifted to the point }(1,1):
\displaystyle \text{(i) }x^2-xy-3y^2-y+2=0
\displaystyle \text{(ii) }xy-y^2-x+y=0
\displaystyle \text{(iii) }xy-x-y+1=0
\displaystyle \text{(iv) }x^2-y^2-2x+2y=0
\displaystyle \text{Answer:}
\displaystyle \text{Since the origin is shifted to }(1,1),\text{ we substitute }x=X+1,\ y=Y+1.
\displaystyle \text{(i) Given }x^2-xy-3y^2-y+2=0
\displaystyle \therefore (X+1)^2-(X+1)(Y+1)-3(Y+1)^2-(Y+1)+2=0
\displaystyle \Rightarrow X^2+2X+1-XY-X-Y-1-3Y^2-6Y-3-Y-1+2=0
\displaystyle \Rightarrow X^2-XY-3Y^2+X-8Y-2=0
\displaystyle \therefore \text{The transformed equation is }X^2-XY-3Y^2+X-8Y-2=0.
\displaystyle \text{(ii) Given }xy-y^2-x+y=0
\displaystyle \therefore (X+1)(Y+1)-(Y+1)^2-(X+1)+(Y+1)=0
\displaystyle \Rightarrow XY+X+Y+1-Y^2-2Y-1-X-1+Y+1=0
\displaystyle \Rightarrow XY-Y^2=0
\displaystyle \therefore \text{The transformed equation is }XY-Y^2=0.
\displaystyle \text{(iii) Given }xy-x-y+1=0
\displaystyle \therefore (X+1)(Y+1)-(X+1)-(Y+1)+1=0
\displaystyle \Rightarrow XY+X+Y+1-X-1-Y-1+1=0
\displaystyle \Rightarrow XY=0
\displaystyle \therefore \text{The transformed equation is }XY=0.
\displaystyle \text{(iv) Given }x^2-y^2-2x+2y=0
\displaystyle \therefore (X+1)^2-(Y+1)^2-2(X+1)+2(Y+1)=0
\displaystyle \Rightarrow X^2+2X+1-Y^2-2Y-1-2X-2+2Y+2=0
\displaystyle \Rightarrow X^2-Y^2=0
\displaystyle \therefore \text{The transformed equation is }X^2-Y^2=0.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the point to which the origin should be shifted after a translation of axes}
\displaystyle \text{so that the following equations contain no first-degree terms:}
\displaystyle \text{(i) }x^2+y^2-4x-8y+3=0
\displaystyle \text{(ii) }x^2+y^2-5x+2y+5=0
\displaystyle \text{(iii) }x^2-12x+4=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the origin be shifted to }(h,k).
\displaystyle \therefore x=X+h,\qquad y=Y+k.
\displaystyle \text{Substituting in }x^2+y^2-4x-8y+3=0,\text{ we get}
\displaystyle (X+h)^2+(Y+k)^2-4(X+h)-8(Y+k)+3=0
\displaystyle \Rightarrow X^2+Y^2+X(2h-4)+Y(2k-8)
\displaystyle \qquad+h^2+k^2-4h-8k+3=0
\displaystyle \text{For the equation to contain no first-degree terms,}
\displaystyle 2h-4=0\qquad\text{and}\qquad 2k-8=0.
\displaystyle \Rightarrow h=2\qquad\text{and}\qquad k=4
\displaystyle \therefore \text{The origin should be shifted to }(2,4).
\displaystyle \text{The transformed equation is }X^2+Y^2-17=0.
\displaystyle \text{(ii) Let the origin be shifted to }(h,k).
\displaystyle \therefore x=X+h,\qquad y=Y+k.
\displaystyle \text{Substituting in }x^2+y^2-5x+2y+5=0,\text{ we get}
\displaystyle (X+h)^2+(Y+k)^2-5(X+h)+2(Y+k)+5=0
\displaystyle \Rightarrow X^2+Y^2+X(2h-5)+Y(2k+2)
\displaystyle \qquad+h^2+k^2-5h+2k+5=0
\displaystyle \text{For the equation to contain no first-degree terms,}
\displaystyle 2h-5=0\qquad\text{and}\qquad 2k+2=0.
\displaystyle \Rightarrow h=\frac{5}{2}\qquad\text{and}\qquad k=-1
\displaystyle \therefore \text{The origin should be shifted to }\left(\frac{5}{2},-1\right).
\displaystyle \text{The transformed equation is }X^2+Y^2-\frac{9}{4}=0.
\displaystyle \text{(iii) Let the origin be shifted to }(h,k).
\displaystyle \therefore x=X+h,\qquad y=Y+k.
\displaystyle \text{Substituting in }x^2-12x+4=0,\text{ we get}
\displaystyle (X+h)^2-12(X+h)+4=0
\displaystyle \Rightarrow X^2+X(2h-12)+h^2-12h+4=0
\displaystyle \text{For the equation to contain no first-degree term in }X,
\displaystyle 2h-12=0
\displaystyle \Rightarrow h=6.
\displaystyle \text{Since the original equation does not contain }y,\text{ the value of }k\text{ is arbitrary.}
\displaystyle \therefore \text{The origin may be shifted to any point }(6,k),\text{ where }k\in\mathbb{R}.
\displaystyle \text{The transformed equation is }X^2-32=0.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Verify that the area of the triangle with vertices }(4,5),\ (7,10)\text{ and }(1,-2)
\displaystyle \text{remains invariant under translation of axes when the origin is shifted to }(-2,1).
\displaystyle \text{Answer:}
\displaystyle \text{Let the vertices of }\triangle ABC\text{ be }A(4,5),\ B(7,10)\text{ and }C(1,-2).
\displaystyle \text{Area of }\triangle ABC
\displaystyle =\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|
\displaystyle =\frac{1}{2}\left|4(10+2)+7(-2-5)+1(5-10)\right|
\displaystyle =\frac{1}{2}|48-49-5|
\displaystyle =\frac{1}{2}|-6|=3\text{ square units.}
\displaystyle \text{Now, the origin is shifted to }(-2,1).
\displaystyle \therefore x=X-2,\qquad y=Y+1,
\displaystyle \text{or }X=x+2,\qquad Y=y-1.
\displaystyle \therefore A'(4+2,5-1)=(6,4)
\displaystyle B'(7+2,10-1)=(9,9)
\displaystyle C'(1+2,-2-1)=(3,-3)
\displaystyle \text{Area of }\triangle A'B'C'
\displaystyle =\frac{1}{2}\left|6(9+3)+9(-3-4)+3(4-9)\right|
\displaystyle =\frac{1}{2}|72-63-15|
\displaystyle =\frac{1}{2}|-6|=3\text{ square units.}
\displaystyle \therefore \text{The area of the triangle remains invariant under the translation of axes.}
\displaystyle \\

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