1

### 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.
2

### 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
3

### 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
4

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

If a straight line passing through the point P(–3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is :
A
x – y + 7 = 0
B
4x – 3y + 24 = 0
C
4x + 3y = 0
D
3x – 4y + 25 = 0

## Explanation

Let the line be ${x \over a} + {y \over b} = 1$

($-$ 3, 4) = $\left( {{a \over 2},{b \over 2}} \right)$

a = $-$6, b = 8

equation of line is 4x $-$ 3y + 24 = 0