Class 11: Cartesian System of Rectangular Co-ordinates – Exercise 22.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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 $(- 2,1)$ .