1
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$$\sqrt 6 $$
2
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
3
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
4
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

Let k be an integer such that the triangle with vertices (k, – 3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point:
A
$$\left( {1,{3 \over 4}} \right)$$
B
$$\left( {1, - {3 \over 4}} \right)$$
C
$$\left( {2,{1 \over 2}} \right)$$
D
$$\left( {2, - {1 \over 2}} \right)$$

Explanation

Given, vertices of triangle are (k, – 3k), (5, k) and (–k, 2).

$${1 \over 2}\left| {\matrix{ k & { - 3k} & 1 \cr 5 & k & 1 \cr { - k} & 2 & 1 \cr } } \right| = \pm 28$$

$$ \Rightarrow $$ k(k - 2) + 3k(5 + k) + 1(10 + k2) = $$ \pm $$ 56

$$ \Rightarrow $$ 5k2 + 13k + 10 = $$ \pm $$ 56

$$ \Rightarrow $$ 5k2 + 13k - 66 = 0

$$ \Rightarrow $$ k = $${{ - 13 \pm \sqrt { - 1151} } \over {10}}$$

So no real solution exist.

or 5k2 + 13k - 46 = 0

$$ \therefore $$ k = $${{ - 23} \over 5}$$ or k = 2

since k is an integer $$ \therefore $$ k = 2

Thus, the coordinate of vertices of triangle are

A(2, -6), B(5, 2) and C(-2, 2).



Now, equation of altitude from vertex A is

y - (-6) = $${{ - 1} \over {\left( {{{2 - 2} \over { - 2 - 5}}} \right)}}\left( {x - 2} \right)$$

$$ \Rightarrow $$ x = 2 .......(1)

Equation of altitude from vertex B is

y - 2 = $${{ - 1} \over {\left( {{{2 + 6} \over { - 2 - 2}}} \right)}}\left( {x - 5} \right)$$

$$ \Rightarrow $$ 2y - 4 = x - 5

$$ \Rightarrow $$ x - 2y = 1 .......(2)

Point H($$\alpha $$, $$\beta $$) lies on both (1) and (2),

$$ \therefore $$ $$\alpha $$ = 2 .........(3)

$$\alpha $$ - 2$$\beta $$ = 1 ......(4)

Solving (3) and (4), we get

$$\alpha $$ = 2 , $$\beta $$ = $${1 \over 2}$$

$$ \therefore $$ Orthocentre is $$\left( {2,{1 \over 2}} \right)$$.

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