1

### 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
2

### 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 $\left( {2,{1 \over 2}} \right)$.
3

### JEE Main 2017 (Online) 9th April Morning Slot

A square, of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle 30o with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is :
A
$2\sqrt 3 - 1$
B
$2\sqrt 3 - 2$
C
$\sqrt 3 - 2$
D
$\sqrt 3 - 1$

## Explanation

Let, coordinate of point A = (x, y).

$\therefore\,\,\,$ For point A,

${x \over {\cos {{30}^ \circ }}}$ = ${y \over {\sin {{30}^ \circ }}}$ = 2

$\Rightarrow$ x = $\sqrt 3$

and y = 1

Similarly, For point B,

${x \over {\cos {{75}^ \circ }}}$ = ${y \over {\sin {{75}^ \circ }}}$ = 2$\sqrt 2$

$\therefore\,\,\,$ x = $\sqrt 3 - 1$

y = $\sqrt 3 + 1$

For point C,

${x \over {cos{{120}^ \circ }}}$ = ${y \over {sin{{120}^ \circ }}}$ = 2

$\Rightarrow$$\,\,\,$ x = $-$1

y = $\sqrt 3$

$\therefore\,\,\,$ Sum of the x - coordinate of the vertices

= 0 + $\sqrt 3$ + $\sqrt 3$ $-$ 1 + ($-$ 1) = 2$\sqrt 3$ $-$ 2
4

### JEE Main 2018 (Offline)

A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is :
A
3x + 2y = 6xy
B
3x + 2y = 6
C
2x + 3y = xy
D
3x + 2y = xy

## Explanation

Let coordinate of point R = (h, k).

Equation of line PQ,

(y $-$ 3) = m (x $-$ 2).

Put y = 0 to get coordinate of point p,

0 $-$ 3 = (x $-$ 2)

$\Rightarrow$ x = 2 $-$ ${3 \over m}$

$\therefore\,\,\,$ p = (2 $-$ ${3 \over m}$, 0)

As p = (h, 0) then

h = 2 $-$ ${3 \over m}$

$\Rightarrow$ ${3 \over m}$ = 2 $-$ h

$\Rightarrow$ m = ${3 \over {2 - h}}$ . . . . . . (1)

Put x = 0 to get coordinate of point Q,

y $-$ 3 $=$ m (0 $-$ 2)

$\Rightarrow$ y = 3 $-$ 2m

$\therefore\,\,\,$ point Q = (0, 3 $-$ 2m)

And From the graph you can see Q = (0, k).

$\therefore\,\,\,$ k = 3 $-$ 2m

$\Rightarrow$ m = ${{3 - k} \over 2}$ . . . . (2)

By comparing (1) and (2) get

${3 \over {2 - h}} = {{3 - k} \over 2}$

$\Rightarrow$ (2 $-$ h)(3 $-$ k) = 6

$\Rightarrow$ 6 $-$ 3h $-$ 2K + hk = 6

$\Rightarrow$ 3 h + 2K = hk

$\therefore\,\,\,$ locus of point R is 3x + 2y= xy