1

### JEE Main 2018 (Online) 15th April Evening Slot

The foot of the perpendicular drawn from the origin, on the line, 3x + y = $\lambda$ ($\lambda$ $\ne$ 0) is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is :
A
1 : 3
B
3 : 1
C
1 : 9
D
9 : 1

## Explanation

Equation of the line, which is perpendicular to the line,

3x + y = $\lambda$($\lambda$ $\ne$0) and passing through origin ,

is given by ${{x - 0} \over 3} = {{y - 0} \over 1} = r$

For foot of perpendicular

r = ${{ - \left( {\left( {3 \times 0} \right) + \left( {1 \times 0} \right) - \lambda } \right)} \over {{3^2} + {1^2}}}$ = ${\lambda \over {10}}$

So, foot of perpendicular P = $\left( {{{3\lambda } \over {10}},{\lambda \over {10}}} \right)$

Given the line meets X-axis where y = 0, so 3x + 0 = $\lambda$

$\Rightarrow$ x = ${\lambda \over 3}$

Hence, coordinates of A = $\left( {{\lambda \over 3},0} \right)$ and meets

Y-axis at B = (0, $\lambda$)

So, BP = $\sqrt {{{\left( {{{3\lambda } \over {10}}} \right)}^2} + {{\left( {{\lambda \over {10}} - \lambda } \right)}^2}}$

$\Rightarrow$   BP = $\sqrt {{{9{\lambda ^2}} \over {100}} + {{81{\lambda ^2}} \over {100}}}$

= BP = $\sqrt {{{90{\lambda ^2}} \over {100}}}$

Now, PA = $\sqrt {{{\left( {{\lambda \over 3} - {{3\lambda } \over {10}}} \right)}^2} + {{\left( {0 - {\lambda \over {10}}} \right)}^2}}$

$\Rightarrow$$\,\,\,$ PA = $\sqrt {{{{\lambda ^2}} \over {900}} + {{{\lambda ^2}} \over {100}}} \Rightarrow PA$ = $\sqrt {{{10{\lambda ^2}} \over {900}}}$

Therefore BP : PA = 9 : 1

2

### 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 $\left( {{3 \over 4},{1 \over 2}} \right)$
3

### 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
4

### 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

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