1

### JEE Main 2016 (Online) 9th April Morning Slot

If a variable line drawn through the intersection of the lines ${x \over 3} + {y \over 4} = 1$ and ${x \over 4} + {y \over 3} = 1,$ meets the coordinate axes at A and B, (A $\ne$ B), then the locus of the midpoint of AB is :
A
6xy = 7(x + y)
B
4(x + y)2 − 28(x + y) + 49 = 0
C
7xy = 6(x + y)
D
14(x + y)2 − 97(x + y) + 168 = 0

## Explanation

L1 : 4x + 3y $-$ 12 = 0

L2 : 3x + 4y $-$ 12 = 0

Equation of line passing through the intersection of these two lines L1 and L2 is

L1 + $\lambda$L2 = 0

$\Rightarrow $$\,\,\,(4x + 3y - 12) + \lambda (3x + 4y - 12) = 0 \Rightarrow$$\,\,\,$ x(4 + 3$\lambda$) + y(3 + 4$\lambda$) $-$ 12(1 + $\lambda$) = 0

this line meets x coordinate at point A and y coordinate at point B.

$\therefore\,\,\,$ Point A = $\left( {{{12\left( {1 + \lambda } \right)} \over {4 + 3\lambda }},0} \right)$

and Point B = $\left( {0,\,\,{{12\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}} \right)$

Let coordinate of midpoint of line AB is (h, k).

$\therefore\,\,\,$ h = ${{6\left( {1 + \lambda } \right)} \over {4 + 3\lambda }}$ . . . . . (1)

and k = ${{6\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}$ . . . . (2)

Eliminate $\lambda$ from (1) and (2), then we get

6(h + k) = 7 hk

$\therefore\,\,\,$ Locus of midpoint of line AB is ,

6(x + y) = 7xy
2

### JEE Main 2016 (Online) 9th April Morning Slot

The point (2, 1) is translated parallel to the line L : x− y = 4 by $2\sqrt 3$ units. If the newpoint Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is :
A
x + y = 2 $-$ $\sqrt 6$
B
x + y = 3 $-$ 3$\sqrt 6$
C
x + y = 3 $-$ 2$\sqrt 6$
D
2x + 2y = 1 $-$ $\sqrt 6$

## Explanation

x $-$ y = 4

To find equation of R

slope of L = 0 is 1

$\Rightarrow$   slope of QR = $-$ 1

Let QR is y = mx + c

y = $-$ x + c

x + y $-$ c = 0

distance of QR from (2, 1) is 2$\sqrt 3$

2$\sqrt 3$ = ${{\left| {2 + 1 - c} \right|} \over {\sqrt 2 }}$ 2$\sqrt 6$ = $\left| {3 - c} \right|$

c $-$ 3 = $\pm 2\sqrt 6$ c = 3 $\pm$ 2$\sqrt 6$

Line can be x + y = 3 $\pm$ 2$\sqrt 6$

x + y = 3 $-$ 2$\sqrt 6$
3

### JEE Main 2016 (Online) 10th April Morning Slot

A ray of light is incident along a line which meets another line, 7x − y + 1 = 0, at the point (0, 1). The ray is then reflected from this point along the line, y + 2x = 1. Then the equation of the line of incidence of the ray of light is :
A
41x − 38y + 38 = 0
B
41x + 25y − 25 = 0
C
41x + 38y − 38 = 0
D
41x − 25y + 25 = 0

## Explanation

Let slope of incident ray be m.

$\therefore$   angle of incidence = angle of reflection $\therefore$   $\left| {{{m - 7} \over {1 + 7m}}} \right| = \left| {{{ - 2 - 7} \over {1 - 14}}} \right| = {9 \over {13}}$

$\Rightarrow$   ${{m - 7} \over {1 + 7m}} = {9 \over {13}}$

or  ${{m - 7} \over {1 + 7m}} = - {9 \over {13}}$

$\Rightarrow$   13m $-$ 91 $=$ 9 + 63m

or   13m $-$ 91 $=$ $-$ 9 $-$ 63m

$\Rightarrow$   50m $=$ $-$ 100  or  76m $=$ 82

$\Rightarrow$   m $=$ $- {1 \over 2}$

or  m $=$ ${{41} \over {38}}$

$\Rightarrow$   y $-$ 1 $=$ $-$ ${1 \over 2}$ (x $-$ 0)

or   y $-$ 1 $=$ ${{41} \over {38}}$ (x $-$ 0)

i.e   x + 2y $-$ 2 $=$ 0

or   38y $-$ 38 $-$ 41x $=$ 0

$\Rightarrow$   41x $-$ 38y + 38 $=$ 0
4

### JEE Main 2016 (Online) 10th April Morning Slot

A straight line through origin O meets the lines 3y = 10 − 4x and 8x + 6y + 5 = 0 at points A and B respectively. Then O divides the segment AB in the ratio :
A
2 : 3
B
1 : 2
C
4 : 1
D
3 : 4

## Explanation

The lines 4x + 3y $-$ 10 = 0 and

8x + 6y + 5 = 0 , are parallel as

${4 \over 8}$  =  ${3 \over 6}$

Now length of perpendicular from

(0, 0, 0) to 4x + 3y $-$ 10 = 0 is,

P1   =   $\left| {{{4\left( 0 \right) + 3\left( 0 \right) - 10} \over {\sqrt {{4^2} + {3^2}} }}} \right|$  =  ${{10} \over 5}$  =  2

Length of perpendicular from

0 (0, 0) to 8x + 6y + 5 = 0 is

P2   =  $\left| {{{8\left( 0 \right) + 6\left( 0 \right) + 5} \over {\sqrt {{6^2} + {8^2}} }}} \right|$   =  ${5 \over {10}}$   =  ${1 \over 2}$

$\therefore\,\,\,$ P1 : P2   =   2 : ${1 \over 2}$   =   4 : 1