 ### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2016 (Offline)

Two sides of a rhombus are along the lines, $$x - y + 1 = 0$$ and $$7x - y - 5 = 0$$. If its diagonals intersect at $$(-1, -2)$$, then which one of the following is a vertex of this rhombus?
A
$$\left( {{{ 1} \over 3}, - {8 \over 3}} \right)$$
B
$$\left( - {{{ 10} \over 3}, - {7 \over 3}} \right)$$
C
$$\left( { - 3, - 9} \right)$$
D
$$\left( { - 3, - 8} \right)$$

## Explanation Let other two sides of rhombus are

$$x - y + \lambda = 0$$

and $$7x - y + \mu = 0$$

then $$O$$ is equidistant from $$AB$$ and $$DC$$ and from $$AD$$ and $$BC$$

$$\therefore$$ $$\left| { - 1 + 2 + 1} \right| = \left| { - 1 + 2 + \lambda } \right| \Rightarrow \lambda = - 3$$

and $$\left| { - 7 + 2 - 5} \right| = \left| { - 7 + 2 + \mu } \right| \Rightarrow \mu = 15$$

$$\therefore$$ Other two sides are $$x-y-3=0$$ and $$7x-y+15=0$$

On solving the equations of sides pairwise, we get

the vertices as $$\left( {{1 \over 3},{{ - 8} \over 3}} \right),\left( {1,2} \right),\left( {{{ - 7} \over 3},{{ - 4} \over 3}} \right),\left( { - 3, - 6} \right)$$
2

### JEE Main 2015 (Offline)

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $$(0, 0)$$ $$(0, 41)$$ and $$(41, 0)$$ is :
A
820
B
780
C
901
D
861

## Explanation The number of integral points lie inside the triangle are

1. If x = 1, then y may be 1, 2, 3, ....., 39

2. If x = 2, then y may be 1, 2, 3, ....., 38

3. If x = 3, then y may be 1, 2, 3, ....., 37

$$\vdots$$

39. If x = 39, then the value of y is 1.

Hence, the number of interior points are

$$1 + 2 + 3 + .... + 39 = {{39 \times 40} \over 2} = 780$$

3

### JEE Main 2014 (Offline)

Let $$a, b, c$$ and $$d$$ be non-zero numbers. If the point of intersection of the lines $$4ax + 2ay + c = 0$$ and $$5bx + 2by + d = 0$$ lies in the fourth quadrant and is equidistant from the two axes then
A
$$3bc - 2ad = 0$$
B
$$3bc + 2ad = 0$$
C
$$2bc - 3ad = 0$$
D
$$2bc + 3ad = 0$$

## Explanation

Since the point of intersection lies on fourth quadrant and equidistant from the two axes,

i.e., let the point be (k, $$-$$k) and this point satisfies the two equations of the given lines.

$$\therefore$$ 4ak $$-$$ 2ak + c = 0 ......... (1)

and 5bk $$-$$ 2bk + d = 0 ..... (2)

From (1) we get, $$k = {{ - c} \over {2a}}$$

Putting the value of k in (2) we get,

$$5b\left( { - {c \over {2a}}} \right) - 2b\left( { - {c \over {2a}}} \right) + d = 0$$

or, $$- {{5bc} \over {2a}} + {{2bc} \over {2a}} + d = 0$$ or, $$- {{3bc} \over {2a}} + d = 0$$

or, $$- 3bc + 2ad = 0$$ or, $$3bc - 2ad = 0$$

4

### 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

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

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

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

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