1

### JEE Main 2019 (Online) 10th January Evening Slot

Two sides of a parallelogram are along the lines, x + y = 3 & x – y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is -
A
(2, 1)
B
(2, 6)
C
(3, 5)
D
(3, 6)

## Explanation

Solving

$\matrix{ {x + y = 3} \cr {x - y = - 3} \cr } \,\, > \,\,A\left( {0,3} \right)$

and  ${{{x_1} + 0} \over 2} = 2;\,\,{x_i} = 4$

similarly y1 = 5

C $\Rightarrow$  (4, 5)

Now equation of BC is x $-$ y = $-$ 1

and equation of CD is x + y = 9

Solving x + y = 9 and x $-$ y = $-$ 3

Point D is (3, 6)
2

### JEE Main 2019 (Online) 10th January Evening Slot

Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant
A
third
B
fourth
C
second
D
first

## Explanation

mBD $\times$ mAD = $-$ 1

$\Rightarrow$  $\left( {{{3 - 2} \over {4 - 0}}} \right) \times \left( {{{b - 0} \over {a - 0}}} \right) = - 1$

$\Rightarrow$  b + 4a = 0      . . . . (i)

mAB $\times$ mCF = $-$ 1

$\Rightarrow$  $\left( {{{\left( {b - 2} \right)} \over {a - 0}}} \right) \times \left( {{3 \over 4}} \right) = - 1$

$\Rightarrow$  3b $-$ 6 = $-$ 4a

$\Rightarrow$  4a + 3b = 6      . . . . .(ii)

From (i) and (ii)

a = ${{ - 3} \over 4}$, b = 3

$\therefore$  IInd quadrant.
3

### JEE Main 2019 (Online) 11th January Evening Slot

If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is :
A
5x + 3y – 11 = 0
B
5x – 3y + 1 = 0
C
3x – 5y + 7 = 0
D
3x + 5y – 13 = 0

## Explanation

co-ordinates of point D are (4, 7)

$\Rightarrow$  line AD is 5x $-$ 3y + 1 = 0
4

### JEE Main 2019 (Online) 12th January Morning Slot

If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, $\beta$), then $\beta$ equals :
A
${{35} \over 3}$
B
$-$ 5
C
$-$ ${{35} \over 3}$
D
5

## Explanation

${{17 - \beta } \over { - 8}} \times {2 \over 3} = - 1$

$\beta$ = 5

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