1

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

If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r cannot be equal to -
A
${7 \over 4}$
B
${5 \over 4}$
C
${3 \over 4}$
D
${3 \over 2}$

## Explanation

r = 1 is obviously true.

Let 0 < r < 1

$\Rightarrow$  r + r2 > 1

$\Rightarrow$  r2 + r $-$ 1 > 0

$\left( {r - {{ - 1 - \sqrt 5 } \over 2}} \right)\left( {r - \left( {{{ - 1 + \sqrt 5 } \over 2}} \right)} \right)$

$\Rightarrow r - {{ - 1 - \sqrt 5 } \over 2}$  or  $r > {{ - 1 + \sqrt 5 } \over 2}$

$r \in \left( {{{\sqrt 5 - 1} \over 2},1} \right)$

${{\sqrt 5 - 1} \over 2} < r < 1$

When r > 1

$\Rightarrow {{\sqrt 5 + 1} \over 2} > {1 \over r} > 1$

$\Rightarrow r \in \left( {{{\sqrt 5 - 1} \over 2},{{\sqrt 5 + 1} \over 2}} \right)$

Now check options
2

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

A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R (3, – 2) are fixed points, then the locus of the centroid of $\Delta$PQR is a line -
A
parallel to y-axis
B
with slope ${2 \over 3}$
C
parallel to x-axis
D
with slope ${3 \over 2}$

## Explanation

Let the centroid of $\Delta$PQR is (h, k) & P is ($\alpha$, $\beta$), then

${{\alpha + 1 + 3} \over 3} = h\,$   and   ${{\beta + 4 - 2} \over 3} = k$

$\alpha = \left( {3h - 4} \right)$   $\beta = \left( {3k - 4} \right)$

Point P($\alpha$, $\beta$) lies on the line 2x $-$ 3y + 4 = 0

$\therefore$  2(3h $-$ 4) $-$ 3 (3k $-$ 2) + 4 = 0

$\Rightarrow$  locus is 6x $-$ 9y + 2 = 0
3

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

If the line 3x + 4y – 24 = 0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is -
A
(3, 4)
B
(2, 2)
C
(4, 4)
D
(4, 3)

## Explanation $\left| {{{3r + 4r - 24} \over 5}} \right| = r$

$7r - 24 = \pm 5r$

$2r = 24$  or  $12r + 24$

$r = 14,\,\,\,r = 2$

then incentre is $(2,2)$
4

### 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)