2 Marks Questions

Question 12 Marks

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

Answer

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

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Question 22 Marks

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

Answer

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

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

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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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