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1

### AIEEE 2004

The equation of the straight line passing through the point $$(4, 3)$$ and making intercepts on the co-ordinate axes whose sum is $$-1$$ is
A
$${x \over 2} - {y \over 3} = 1$$ and $${x \over -2} +{y \over 1} = 1$$
B
$${x \over 2} - {y \over 3} = -1$$ and $${x \over -2} +{y \over 1} = -1$$
C
$${x \over 2} + {y \over 3} = 1$$ and $${x \over 2} +{y \over 1} = 1$$
D
$${x \over 2} + {y \over 3} = -1$$ and $${x \over -2} +{y \over 1} = -1$$

## Explanation

Let the required line be $${x \over a} + {y \over b} = 1.......\left( 1 \right)$$

then $$a+b=-1$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.........\left( 2 \right)$$

$$(1)$$ passes through $$(4,3),$$ $$\Rightarrow {4 \over a} + {3 \over b} = 1$$

$$\Rightarrow 4b + 3a = ab\,\,...............\left( 3 \right)$$

Eliminating $$b$$ from $$(2)$$ and $$(3),$$ we get

$${a^2} - 4 = 0 \Rightarrow a = \pm 2 \Rightarrow b = - 3$$

$$1$$

$$\therefore$$ Equation of straight lines are

$${x \over 2} + {y \over { - 3}} = 1$$

or $${x \over { - 2}} + {y \over 1} = 1$$
2

### 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

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

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

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

$$c = {1 \over 2}\left( {{a_2}^2 + {b_2}^2 - a{{{}_1}^2} - {b_1}^2} \right)$$
3

### 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.
4

### 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

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

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

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

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

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