 ### JEE Mains Previous Years Questions with Solutions

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1

### 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

2

### AIEEE 2011

The lines $${L_1}:y - x = 0$$ and $${L_2}:2x + y = 0$$ intersect the line $${L_3}:y + 2 = 0$$ at $$P$$ and $$Q$$ respectively. The bisector of the acute angle between $${L_1}$$ and $${L_2}$$ intersects $${L_3}$$ at $$R$$.

Statement-1: The ratio $$PR$$ : $$RQ$$ equals $$2\sqrt 2 :\sqrt 5$$
Statement-2: In any triangle, bisector of an angle divide the triangle into two similar triangles.

A
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
B
Statement-1 is true, Statement-2 is false.
C
Statement-1 is false, Statement-2 is true.
D
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

## Explanation $${L_1}:y - x = 0$$

$${L_2}:2x + y = 0$$

$${L_3}:y + 2 = 0$$

On solving the equation of line $${L_1}$$ and $${L_2}$$ we get their point of

intersection $$(0, 0)$$ i.e., origin $$O.$$

On solving the equation of line $${L_1}$$ and $${L_3},$$

we get $$P=(-2, -2).$$

Similarly, we get $$Q = \left( { - 1, - 2} \right)$$

We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [ Angle Bisector Theorem of a Triangle ]

$$\therefore$$ $${{PR} \over {RQ}} = {{OP} \over {OQ}} = {{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}} } \over {\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} }}$$

$$= {{2\sqrt 2 } \over {\sqrt 5 }}$$
3

### AIEEE 2010

The line $$L$$ given by $${x \over 5} + {y \over b} = 1$$ passes through the point $$\left( {13,32} \right)$$. The line K is parrallel to $$L$$ and has the equation $${x \over c} + {y \over 3} = 1.$$ Then the distance between $$L$$ and $$K$$ is
A
$$\sqrt {17}$$
B
$${{17} \over {\sqrt {15} }}$$
C
$${{23} \over {\sqrt {17} }}$$
D
$${{23} \over {\sqrt {15} }}$$

## Explanation

Slope of line $$L = - {b \over 5}$$

Slope of line $$K = - {3 \over c}$$

Line $$L$$ is parallel to line $$k.$$

$$\Rightarrow {b \over 5} = {3 \over c} \Rightarrow bc = 15$$

$$(13,32)$$ is a point on $$L.$$

$$\therefore$$ $${{13} \over 5} + {{32} \over b} = 1 \Rightarrow {{32} \over b} = - {8 \over 5}$$

$$\Rightarrow b = - 20 \Rightarrow c = - {3 \over 4}$$

Equation of $$K:$$ $$y - 4x = 3$$

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

Distance between $$L$$ and $$K$$

$$= {{\left| {52 - 32 + 3} \right|} \over {\sqrt {17} }} = {{23} \over {\sqrt {17} }}$$
4

### AIEEE 2009

Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point $$(1, 0)$$ to the distance from the point $$(-1, 0)$$ is equal to $${1 \over 3}$$. Then the circumcentre of the triangle ABC is at the point:
A
$$\left( {{5 \over 4},0} \right)$$
B
$$\left( {{5 \over 2},0} \right)$$
C
$$\left( {{5 \over 3},0} \right)$$
D
$$\left( {0,0} \right)$$

## Explanation

Given that

$$P\left( {1,0} \right),Q\left( { - 1,0} \right)$$

and $${{AP} \over {AQ}} = {{BP} \over {BQ}} = {{CP} \over {CQ}} = {1 \over 3}$$

$$\Rightarrow 3AP = AQ$$

$$\,\,\,\,\,\,$$ Let $$A = (x,y)$$ then $$3AP = AQ \Rightarrow 9A{P^2} = A{Q^2}$$

$$\Rightarrow 9{\left( {x - 1} \right)^2} + 9{y^2} = {\left( {x + 1} \right)^2} + y{}^2$$

$$\Rightarrow 9{x^2} - 18x + 9 + 9{y^2} = {x^2} + 2x + 1 + {y^2}$$

$$\Rightarrow 8{x^2} - 20x + 8{y^2} + 8 = 0$$

$$\Rightarrow {x^2} + {y^2} - {5 \over 3}x + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$\therefore$$ A lies on the circle given by eq. $$(1).$$ As $$B$$ and $$C$$

also follow the same condition, - they must lie on the same circle.

$$\therefore$$ Center of circumcircle of $$\Delta ABC$$

$$=$$ Center of circle given by $$\left( 1 \right) = \left( {{5 \over 4},0} \right)$$

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