1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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