1

### JEE Main 2019 (Online) 9th January Morning Slot

Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statements is true ?
A
The lines are not concurrent
B
The lines are concurrent at the point $\left( {{3 \over 4},{1 \over 2}} \right)$
C
The lines are all parallel
D
Each line passes through the origin

## Explanation

Equation of lines;

px + qy + r = 0   . . . . . (1)

Also given

3p + 2q + 4r = 0   . . . . . . (2)

divide equation (2) by 4, we get

${3 \over 4}P + {2 \over 4}q + r = 0$   . . . . (3)

By comparing (1) and (3) we get,

x = ${3 \over 4}$ and y = ${2 \over 4}$ = ${1 \over 2}$

For any value of p,q and r, the equation of set of lines will pan through $\left( {{3 \over 4},{1 \over 2}} \right)$
2

### JEE Main 2019 (Online) 9th January Evening Slot

Let the equations of two sides of a triangle be 3x $-$ 2y + 6 = 0 and 4x + 5y $-$ 20 = 0. If the orthocentre of this triangle is at (1, 1), then the equation of its third side is :
A
122y $-$ 26x $-$ 1675 = 0
B
122y + 26x + 1675 = 0
C
26x + 61y + 1675 = 0
D
26x $-$ 122y $-$ 1675 = 0

## Explanation

4x + 5y $-$ 20 = 0       . . .(1)

3x $-$ 2y + 6 = 0       . . . (2)

orthocentre is (1, 1)

line perpendicular to 4x + 5y $-$ 20 = 0

and passes through (1, 1) is

(y $-$ 1) = ${5 \over 4}$(x $-$ 1)

$\Rightarrow$  5x $-$ 4y = 1       . . .(3)

and line $\bot$ to 3x $-$ 2y + 6 = 0

and passes through (1, 1)

y $-$ 1 = $-$ ${2 \over 3}$ (x $-$ 1)

$\Rightarrow$  2x + 3y = 5       . . .(4)

Solving (1) and (4) we get C$\left( {{{35} \over 2}, - 10} \right)$

Solving (2) and (3) we get A $\left( { - 13,{{ - 33} \over 2}} \right)$

Side BC is y + 10 = ${{{{ - 33} \over 2} + 10} \over { - 13 - {{35} \over 2}}}\left( {x - {{35} \over 2}} \right)$

$\Rightarrow$  y + 10 = ${{13} \over {61}}\left( {x - {{35} \over 2}} \right)$

$\Rightarrow$  26x $-$ 122y $-$ 1675 = 0
3

### 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
4

### 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