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1

### AIEEE 2005

If a vertex of a triangle is $$(1, 1)$$ and the mid points of two sides through this vertex are $$(-1, 2)$$ and $$(3, 2)$$ then the centroid of the triangle is
A
$$\left( { - 1,{7 \over 3}} \right)$$
B
$$\left( {{{ - 1} \over 3},{7 \over 3}} \right)$$
C
$$\left( { 1,{7 \over 3}} \right)$$
D
$$\left( {{{ 1} \over 3},{7 \over 3}} \right)$$

## Explanation

Vertex of triangle is $$\left( {1,\,1} \right)$$ and midpoint of sides through -

this vertex is $$\left( { - 1,\,2} \right)$$ and $$\left( {3,2} \right)$$

$$\Rightarrow$$ vertex $$B$$ and $$C$$ come out to be $$\left( { - 3,3} \right)$$ and $$\left( {5,3} \right)$$

$$\therefore$$ centroid is $${{1 - 3 + 5} \over 3},{{1 + 3 + 5} \over 3} \Rightarrow \left( {1,{7 \over 3}} \right)$$
2

### 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.
3

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

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

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

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

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