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1

### AIEEE 2008

The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept -4. Then a possible value of k is
A
1
B
2
C
-2
D
-4

## Explanation

Slope of $$PQ = {{3 - 4} \over {k - 1}} = {{ - 1} \over {k - 1}}$$

$$\therefore$$ Slope of perpendicular bisector of

$$PQ = \left( {k - 1} \right)$$

Also mid point of

$$PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).$$

Equation of perpendicular bisector is

$$y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)$$

$$\Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)$$

$$\Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0$$

$$\therefore$$ $$y$$-intercept $$= {{8 - {k^2}} \over { - 2}} = - 4$$

$$\Rightarrow$$ $$8 - {k^2} = - 8$$ or $${k^2} = 16 \Rightarrow k = \pm 4$$
2

### AIEEE 2007

If one of the lines of $$m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0$$ is a bisector of angle between the lines $$xy = 0,$$ then $$m$$ is
A
$$1$$
B
$$2$$
C
$$-1/2$$
D
$$-2$$

## Explanation

Equation of bisectors of lines, $$xy=0$$ are $$y = \pm x$$ $$\therefore$$ Put $$y = \pm \,x$$ in the given equation

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

$$\therefore$$ $$m{x^2} + \left( {1 - {m^2}} \right){x^2} - m{x^2} = 0$$

$$\Rightarrow 1 - {m^2} = 0 \Rightarrow m = \pm 1$$
3

### AIEEE 2007

Let $$P = \left( { - 1,0} \right),\,Q = \left( {0,0} \right)$$ and $$R = \left( {3,3\sqrt 3 } \right)$$ be three point. The equation of the bisector of the angle $$PQR$$ is
A
$${{\sqrt 3 } \over 2}x + y = 0$$
B
$$x + \sqrt {3y} = 0$$
C
$$\sqrt 3 x + y = 0$$
D
$$x + {{\sqrt 3 } \over 2}y = 0$$

## Explanation

Given : The coordinates of points $$P,Q,R$$ are $$(-1,0),$$

$$\left( {0,0} \right),\,\left( {3,3\sqrt 3 } \right)$$ respectively Slope of $$QR$$ $$= {{{y_2} - {y_1}} \over {{x_2} - {x_1}}} = {{3\sqrt 3 } \over 3}$$

$$\Rightarrow \tan \theta = \sqrt 3 \Rightarrow \theta = {\pi \over 3}$$

$$\Rightarrow \angle RQX = {\pi \over 3}$$

$$\therefore$$ $$\angle RQP = \pi - {\pi \over 3} = {{2\pi } \over 3}$$

Let $$QM$$ bisects the $$\angle PQR,$$

$$\therefore$$ Slope of the line $$QM=tan$$ $${{2\pi } \over 3} = - \sqrt 3$$

$$\therefore$$ Equation of line $$QM$$ is $$\left( {y - 0} \right) = - \sqrt 3 \left( {x - 0} \right)$$

$$\Rightarrow y = - \sqrt 3 \,x \Rightarrow \sqrt 3 x + y = 0$$
4

### AIEEE 2007

Let A $$\left( {h,k} \right)$$, B$$\left( {1,1} \right)$$ and C $$(2, 1)$$ be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is $$1$$ square unit, then the set of values which $$'k'$$ can take is given by
A
$$\left\{ { - 1,3} \right\}$$
B
$$\left\{ { - 3, - 2} \right\}$$
C
$$\left\{ { 1,3} \right\}$$
D
$$\left\{ {0,2} \right\}$$

## Explanation

Given : The vertices of a right angled triangle $$A\left( {1,k} \right),$$

$$B\left( {1,1} \right)$$ and $$C\left( {2,1} \right)$$ and area of $$\Delta ABC = 1$$ square unit We know that, area of night angled triangle

$$= {1 \over 2} \times BC \times AB = 1 = {1 \over 2}\left( 1 \right)\left| {\left( {k - 1} \right)} \right|$$

$$\Rightarrow \pm \left( {k - 1} \right) = 2 \Rightarrow k = - 1,3$$

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