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1

### AIEEE 2005

The line parallel to the $$x$$ - axis and passing through the intersection of the lines $$ax + 2by + 3b = 0$$ and $$bx - 2ay - 3a = 0,$$ where $$(a, b)$$ $$\ne$$ $$(0, 0)$$ is
A
below the $$x$$ - axis at a distance of $${3 \over 2}$$ from it
B
below the $$x$$ - axis at a distance of $${2 \over 3}$$ from it
C
above the $$x$$ - axis at a distance of $${3 \over 2}$$ from it
D
above the $$x$$ - axis at a distance of $${2 \over 3}$$ from it

## Explanation

The line passing through the intersection of lines

$$ax + 2by = 3b = 0$$

and $$bx - 2ay - 3a = 0$$ is

$$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$$

$$\Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$$

As this line is parallel to $$x$$-axis.

$$\therefore$$ $$a + b\lambda = 0 \Rightarrow \lambda = - a/b$$

$$\Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$$

$$\Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$$

$$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$$

$$y\left( {{{2{b^2} + 2{a^2}} \over b}} \right) = - \left( {{{3{b^2} + 3{a^2}} \over b}} \right)$$

$$y = {{ - 3\left( {{a^2} + {b^2}} \right)} \over {2\left( {{b^2} + {a^2}} \right)}} = {{ - 3} \over 2}$$

So it is $$3/2$$ units below $$x$$-axis.
2

### AIEEE 2004

If one of the lines given by $$6{x^2} - xy + 4c{y^2} = 0$$ is $$3x + 4y = 0,$$ then $$c$$ equals
A
$$-3$$
B
$$-1$$
C
$$3$$
D
$$1$$

## Explanation

$$3x+4y=0$$ is one of the lines of the pair

$$6{x^2} - xy + 4c{y^2} = 0,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$

Put $$\,\,\,\,\,y = - {3 \over 4}x,$$

we get $$6{x^2} + {3 \over 4}{x^2} + 4c{\left( { - {3 \over 4}x} \right)^2} = 0$$

$$\Rightarrow 6 + {3 \over 4} + {{9c} \over 4} = 0 \Rightarrow c = - 3$$
3

### AIEEE 2004

If the sum of the slopes of the lines given by $${x^2} - 2cxy - 7{y^2} = 0$$ is four times their product $$c$$ has the value
A
$$-2$$
B
$$-1$$
C
$$2$$
D
$$1$$

## Explanation

Let the lines be $$y = {m_1}x$$ and $$y = {m_2}x$$ then

$${m_1} + {m_2} = - {{2c} \over 7}$$ and $${m_1}{m_2} = - {1 \over 7}$$

Given $${m_1} + {m_2} = 4m{}_1{m_2}$$

$$\Rightarrow {{2c} \over 7} = - {4 \over 7} \Rightarrow c = 2$$
4

### AIEEE 2004

Let $$A\left( {2, - 3} \right)$$ and $$B\left( {-2, 1} \right)$$ be vertices of a triangle $$ABC$$. If the centroid of this triangle moves on the line $$2x + 3y = 1$$, then the locus of the vertex $$C$$ is the line
A
$$3x - 2y = 3$$
B
$$2x - 3y = 7$$
C
$$3x + 2y = 5$$
D
$$2x + 3y = 9$$

## Explanation

Let the vertex $$C$$ be $$(h,k),$$ then the

centroid of $$\Delta ABC$$ is $$\left( {{{2 + (- 2) + h} \over 3},{{ - 3 + 1 + k} \over 3}} \right)$$

or $$\left( {{h \over 3},{{ - 2 + k} \over 3}} \right).$$ It lies on $$2x+3y=1$$

$$\Rightarrow {{2h} \over 3} - 2 + k = 1$$

$$\Rightarrow 2h + 3k = 9$$

$$\therefore$$ Locus of $$C$$ is $$2x+3y=9$$

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