(Each question 2 marks)

Question 12 Marks

Find what the following equations become when the origin is shifted to the point $(1,1) ? x^2+y^2-2 x-2 y=0$

Answer

We have $x^2+y^2-2 x-2 y=0$
Substituting $x=X+1, y+1$ in the given equation, we get
$(X+1)^2-(Y-1)^2-2(X+1)+2(Y+1)=0$
$\Rightarrow X^2+1+2 X-\left(Y^2+1+2 Y\right)-2 X-2+2 Y+2=0$
$\Rightarrow X^2+1-Y^2-1-2 Y+2 Y=0$
$\Rightarrow X^2-Y^2=0$

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Question 22 Marks

Find what the following equations become when the origin is shifted to the point $(1,1) ? x^2+y^2-x+2 y=0$

Answer

We have, $x^2+y^2-x+2 y=0$
Substituting $x=X+1, y+1$ in the given equation, we get
$(X+1)(Y+1)-(Y+1)-(X+1)-(Y+1)=0$
$\Rightarrow X Y+Y+Y+1-\left(Y^2+1-2 Y\right)-X-1-Y+1=0$
$\Rightarrow X Y+2 Y-Y^2-1-2 Y+1=0$
$\Rightarrow X Y-Y^2=0$

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Question 32 Marks

Find the locus of a point equidistant from the point(2, 4) and the y-axis.

Answer

Let p(h, k) be any point on the locus and let A(2, 4) and B(0, k). Then, $\text{PA}=\text{PB}$ $\Rightarrow \text{PA}^2=\text{PB}^2$ $\Rightarrow \bigg[\sqrt{(2-\text{h})^2+(4-\text{k})^2}\bigg]^2=\bigg[\sqrt{(0-\text{h})^2+(\text{k-k})^2}\bigg]^2$ $\Rightarrow (2-\text{h})^2+(4-\text{h})^2=(0-\text{h})^2+(0)^2$ $\Rightarrow 4+\text{h}^2-4\text{h}+16+\text{k}^2-8\text{k}=\text{h}^2$ $\Rightarrow \text{k}^2-8\text{k}-4\text{h}+20=0$ Hence, locus of (h, k) is $\text{y}^2-8\text{y}-4\text{y}+20=0$ Let P(h, k) be any point on the locus and let AC(2, 4) and B(0, k) be the given points.

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Question 42 Marks

Find the locus of a point which is equidistant from (1, 3) and the x-axis.

Answer

Let P(h, k) be any point on the locus and let A(1, 3) and B(h, 0).Then,$\text{PA}=\text{PB}$
$\Rightarrow \text{P}\text{A}^2=\text{P}\text{B}^2$
$\Rightarrow (1-\text{h})^2+(3-\text{k})^2=(\text{h}-\text{h})^2+(0-\text{k})^2$
$\Rightarrow 1+\text{h}^2-2\text{h}+9+\text{k}^2-6\text{k}=0+\text{k}^2$
$\Rightarrow \text{h}^2-2\text{h}-6\text{k}+10=0$
Hence, locus of (h, k) is $\text{x}^2-2\text{x}-6\text{y}+1=0$

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Question 52 Marks

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

We have, $(x-a)^2+(y-b)^2=r^2 \ldots .$. (i) Substituting $x=x+(a-c), y=y+b$ in the equcation(i), we get $[x+a-c-a]^2+$ $[y+b-b]^2=r^2 \Rightarrow[x-c]^2+[y]^2=r^2 \Rightarrow x^2+c^2-2 x c+y^2=r^2 \Rightarrow x^2+y^2-2 x c=r^2-c^2$ Hence, the required equation is $x^2+y^2-2 x c=r^2-c^2$

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Question 62 Marks

The points A(2, 0), B(9, 1), C(11, 6)and D(4, 4) ar the vertices of a quadrilateral ABCD, Determine whether ABCD is a rhombus or not.

Answer

It is given that A(2, 0), B(9, 1), C(11, 6) and D(4, 4) are the vetioes a quadrilateral. Now, Coordinates of the mid-point of AC are $\bigg(\frac{2+11}{2},\frac{0+6}{2}\bigg)=\bigg(\frac{13}{2},3\bigg)$ Coordinates of the mid-point of BD are $\bigg(\frac{9+4}{2},\frac{1+4}{2}\bigg)=\bigg(\frac{13}{2},\frac{5}{2}\bigg)$ Thus, AC and BD do not have the same mid-point, hence ABCD is not a paralleogram.$\therefore $ ABCD is not a rhombus.

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Question 72 Marks

Find what the following equations become when the origin is shifted to the point (1, 1)? xy − x − y + 1 = 0

Answer

We have,xy − x − y + 1 = 0
Substituting x = X + 1, y + 1 in the given equation, we get
(X + 1)(Y + 1) - (X + 1) - (Y + 1) + 1 = 0
⇒ XY + X + Y + 1 - X - 1 - Y - 1 + 1 = 0
⇒ XY = 0

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MATHS (Each question 2 marks)

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