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1

### AIEEE 2003

If the equation of the locus of a point equidistant from the point $$\left( {{a_{1,}}{b_1}} \right)$$ and $$\left( {{a_{2,}}{b_2}} \right)$$ is
$$\left( {{a_1} - {b_2}} \right)x + \left( {{a_1} - {b_2}} \right)y + c = 0$$ , then the value of $$'c'$$ is
A
$$\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2}$$
B
$${1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)$$
C
$${{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}$$
D
$${1 \over 2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right)$$.

## Explanation

$${\left( {x - {a_1}} \right)^2} + \left( {y - {b_1}} \right){}^2 = \left( {x - {a_2}} \right){}^2 + {\left( {y - {b_2}} \right)^2}$$

$$\left( {a{}_1 - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$+ {1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right) = 0$$

$$c = {1 \over 2}\left( {{a_2}^2 + {b_2}^2 - a{{{}_1}^2} - {b_1}^2} \right)$$
2

### AIEEE 2003

If $${x_1},{x_2},{x_3}$$ and $${y_1},{y_2},{y_3}$$ are both in G.P. with the same common ratio, then the points $$\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$$ and $$\left( {{x_3},{y_3}} \right)$$
A
are vertices of a triangle
B
lie on a straight line
C
lie on an ellipse
D
lie on a circle

## Explanation

Taking co-ordinates as

$$\left( {{x \over r},{y \over r}} \right);\left( {x,y} \right)\,\,\& \,\,\left( {xr,yr} \right)$$

Then slope of line joining

$$\left( {{x \over r},{y \over r}} \right),\left( {x,y} \right) = {{y\left( {1 - {1 \over r}} \right)} \over {x\left( {1 - {1 \over r}} \right)}} = {y \over x}$$

and slope of line joining $$(x,y)$$ and $$(xr, yr)$$

$$= {{y\left( {r - 1} \right)} \over {x\left( {r - 1} \right)}} = {y \over x}$$

$$\therefore$$ $${m_1} = {m_2}$$

$$\Rightarrow$$ Points lie on the straight line.
3

### AIEEE 2003

Locus of centroid of the triangle whose vertices are $$\left( {a\cos t,a\sin t} \right),\left( {b\sin t, - b\cos t} \right)$$ and $$\left( {1,0} \right),$$ where $$t$$ is a parameter, is
A
$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$
B
$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$
C
$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$
D
$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$

## Explanation

$$x = {{a\cos t + b\sin t + 1} \over 3}$$

$$\Rightarrow a\cos t + b\sin t = 3x - 1$$

$$y = {{a\sin t - b\cos t} \over 3}$$

$$\Rightarrow a\sin t - b\cos t = 3y$$
4

### AIEEE 2003

If the pair of straight lines $${x^2} - 2pxy - {y^2} = 0$$ and $${x^2} - 2qxy - {y^2} = 0$$ be such that each pair bisects the angle between the other pair, then
A
$$pq = -1$$
B
$$p = q$$
C
$$p = -q$$
D
$$pq = 1$$.

## Explanation

Equation of bisectors of second pair of straight lines is,

$$q{x^2} + 2xy - q{y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

It must be identical to the first pair

$${x^2} - 2\,pxy - {y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(... 2 \right)$$

from $$(1)$$ and $$(2)$$ $${q \over 1} = {2 \over { - 2p}} = {{ - q} \over { - 1}}$$

$$\Rightarrow pq = - 1.$$

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