 ### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2014 (Offline)

Let $$PS$$ be the median of the triangle with vertices $$P(2, 2)$$, $$Q(6, -1)$$ and $$R(7, 3)$$. The equation of the line passing through $$(1, -1)$$ band parallel to PS is:
A
$$4x + 7y + 3 = 0$$
B
$$2x - 9y - 11 = 0$$
C
$$4x - 7y - 11 = 0$$
D
$$2x + 9y + 7 = 0$$

## Explanation

Let $$P,Q,R,$$ be the vertices of $$\Delta PQR$$ Since $$PS$$ is the median, $$S$$ is mid-point of $$QR$$

So, $$S = \left( {{{7 + 6} \over 2},{{3 - 1} \over 2}} \right) = \left( {{{13} \over 2},1} \right)$$

Now, slope of $$PS$$ $$= {{2 - 1} \over {2 - {{13} \over 2}}} = - {2 \over 9}$$

Since, required line is parallel to $$PS$$ therefore slope of required line $$=$$ slope of $$PS$$

Now, equation of line passing through $$(1, -1)$$ and having slope $$- {2 \over 9}$$ is

$$y - \left( { - 1} \right) = - {2 \over 9}\left( {x - 1} \right)$$

$$9y + 9 = - 2x + 2$$

$$\Rightarrow 2x + 9y + 7 = 0$$
2

### JEE Main 2013 (Offline)

The $$x$$-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as $$(0, 1) (1, 1)$$ and $$(1, 0)$$ is :
A
$$2 + \sqrt 2$$
B
$$2 - \sqrt 2$$
C
$$1 + \sqrt 2$$
D
$$1 - \sqrt 2$$

## Explanation

From the figure, we have

$$a = 2,b = 2\sqrt 2 ,c = 2$$

$${x_1} = 0,\,{x^2} = 0,\,{x_3} = 2$$ Now, $$x$$-co-ordinate of incenter is given as

$${{a{x_1} + b{x_2} + c{x_3}} \over {a + b + c}}$$

$$\Rightarrow x$$-coordinate of incentre

$$= {{2 \times 0 + 2\sqrt 2 .0 + 2.2} \over {2 + 2 + 2\sqrt 2 }}$$

$$=$$ $${2 \over {2 + \sqrt 2 }} = 2 - \sqrt 2$$
3

### JEE Main 2013 (Offline)

A ray of light along $$x + \sqrt 3 y = \sqrt 3$$ gets reflected upon reaching $$X$$-axis, the equation of the reflected ray is
A
$$y = x + \sqrt 3$$
B
$$\sqrt 3 y = x - \sqrt 3$$
C
$$y = \sqrt 3 x - \sqrt 3$$
D
$$\sqrt 3 y = x - 1$$

## Explanation

$$x + \sqrt 3 y = \sqrt 3$$ or $$y = - {1 \over {\sqrt 3 }}x + 1$$ Let $$\theta$$ be the angle which the line makes with the positive x-axis.

$$\therefore$$ $$\tan \theta = - {1 \over {\sqrt 3 }} = \tan \left( {\pi - {\pi \over 6}} \right)$$ or $$\theta = \pi - {\pi \over 6}$$

$$\therefore$$ $$\angle ABC = {\pi \over 6}$$; $$\therefore$$ $$\angle DBE = {\pi \over 6}$$

$$\therefore$$ the equation of the line BD is,

$$y = \tan {\pi \over 6}x + c$$ or $$y = {x \over {\sqrt 3 }} + c$$ ..... (1)

The line $$x + \sqrt 3 y = \sqrt 3$$ intersects the x-axis at $$B(\sqrt 3 ,0)$$ and, the line (1) passes through $$B(\sqrt 3 ,0)$$.

$$\therefore$$ $$0 = {{\sqrt 3 } \over {\sqrt 3 }} + c$$ or, c = $$-$$1

Hence, the equation of the reflected ray is,

$$y = {x \over {\sqrt 3 }} - 1$$ or $$y\sqrt 3 = x - \sqrt 3$$

4

### AIEEE 2012

If the line $$2x + y = k$$ passes through the point which divides the line segment joining the points $$(1, 1)$$ and $$(2, 4)$$ in the ratio $$3 : 2$$, then $$k$$ equals :
A
$${{29 \over 5}}$$
B
$$5$$
C
$$6$$
D
$${{11 \over 5}}$$

## Explanation

The point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2 is

$$= \left( {{{3 \times 2 + 2 \times 1} \over {3 + 2}},{{3 \times 4 + 2 \times 1} \over {3 + 2}}} \right)$$

$$= \left( {{{6 + 2} \over 5},{{12 + 2} \over 5}} \right) = \left( {{8 \over 5},{{14} \over 5}} \right)$$

Since the line 2x + y = k passes through this point,

$$\therefore$$ $$2 \times {8 \over 5} + {{14} \over 5} = k$$ or $${{30} \over 5} = k$$ or, k = 6

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