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1

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

$$\Rightarrow 6 + {3 \over 4} + {{9c} \over 4} = 0 \Rightarrow c = - 3$$
2

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

$$\Rightarrow {{2c} \over 7} = - {4 \over 7} \Rightarrow c = 2$$
3

### 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 $$2x+3y=9$$
4

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

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