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1

### JEE Main 2021 (Online) 27th August Evening Shift

The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m $$-$$ n = 0 and mn + nl + lm = 0, is :
A
$${\pi \over 2}$$
B
$$\pi - {\cos ^{ - 1}}\left( {{4 \over 9}} \right)$$
C
$${\cos ^{ - 1}}\left( {{8 \over 9}} \right)$$
D
$${\pi \over 3}$$

## Explanation

n = 2 (l + m)

lm + n(l + m) = 0

lm + 2(l + m)2 = 0

2l2 + 2m2 + 5ml = 0

$$2{\left( {{l \over m}} \right)^2} + 2 + 5\left( {{l \over m}} \right) = 0$$

2t2 + 5t + 2 = 0

(t + 2)(2t + 1) = 0

$$\Rightarrow t = - 2; - {1 \over 2}$$

(i) $${l \over m} = - 2$$

$${n \over m} = - 2$$

($$-$$2m, m, $$-$$2m)

($$-$$2, 1, $$-$$2)

(ii) $${l \over m} = - {1 \over 2}$$

n = $$-$$2l

(l, $$-$$2l, $$-$$2l)

(1, $$-$$2, $$-$$2)

$$\cos \theta = {{ - 2 - 2 + 4} \over {\sqrt 9 \sqrt 9 }} = 0 \Rightarrow 0 = {\pi \over 2}$$
2

### JEE Main 2021 (Online) 27th August Morning Shift

Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :
A
3x2 $$-$$ 2y $$-$$ 6 = 0
B
3x2 + 2y $$-$$ 6 = 0
C
2x2 + 3y $$-$$ 9 = 0
D
2x2 $$-$$ 3y + 9 = 0

## Explanation

A(0, 6) and B(2t, 0) Perpendicular bisector of AB is

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

So, $$C = \left( {0,3 - {{{t^2}} \over 3}} \right)$$

Let P be (h, k)

$$h = {t \over 2};k = \left( {3 - {{{t^2}} \over 6}} \right)$$

$$\Rightarrow k = 3 - {{4{h^2}} \over 6} \Rightarrow 2{x^2} + 3y - 9 = 0$$
3

### JEE Main 2021 (Online) 26th August Morning Shift

Let ABC be a triangle with A($$-$$3, 1) and $$\angle$$ACB = $$\theta$$, 0 < $$\theta$$ < $${\pi \over 2}$$. If the equation of the median through B is 2x + y $$-$$ 3 = 0 and the equation of angle bisector of C is 7x $$-$$ 4y $$-$$ 1 = 0, then tan$$\theta$$ is equal to :
A
$${1 \over 2}$$
B
$${3 \over 4}$$
C
$${4 \over 3}$$
D
2

## Explanation $$\therefore$$ $$M\left( {{{a - 3} \over 2},{{b + 1} \over 2}} \right)$$ lies on 2x + y $$-$$ 3 = 0

$$\Rightarrow$$ 2a + b = 11 ...........(i)

$$\because$$ C lies on 7x $$-$$ 4y = 1

$$\Rightarrow$$ 7a $$-$$ 4b = 1 ......... (ii)

$$\therefore$$ by (i) and (ii) : a = 3, b = 5

$$\Rightarrow$$ C(3, 5)

$$\therefore$$ mAC = 2/3

Also, mCD = 7/4

$$\Rightarrow \tan {\theta \over 2} = \left| {{{{2 \over 3} - {4 \over 4}} \over {1 + {{14} \over {12}}}}} \right| \Rightarrow \tan {\theta \over 2} = {1 \over 2}$$

$$\Rightarrow \tan \theta = {{2.{1 \over 2}} \over {1 - {1 \over 4}}} = {4 \over 3}$$
4

### JEE Main 2021 (Online) 25th July Evening Shift

Let the equation of the pair of lines, y = px and y = qx, can be written as (y $$-$$ px) (y $$-$$ qx) = 0. Then the equation of the pair of the angle bisectors of the lines x2 $$-$$ 4xy $$-$$ 5y2 = 0 is :
A
x2 $$-$$ 3xy + y2 = 0
B
x2 + 4xy $$-$$ y2 = 0
C
x2 + 3xy $$-$$ y2 = 0
D
x2 $$-$$ 3xy $$-$$ y2 = 0

## Explanation

Equation of angle bisector of homogeneous
equation of pair of straight line ax2 + 2hxy + by2 is

$${{{x^2} - {y^2}} \over {a - b}} = {{xy} \over h}$$

for x2 – 4xy – 5y2 = 0

a = 1, h = – 2, b = – 5

So, equation of angle bisector is

$${{{x^2} - {y^2}} \over {1 - ( - 5)}} = {{xy} \over { - 2}}$$

$${{{x^2} - {y^2}} \over 6} = {{xy} \over { - 2}}$$

$$\Rightarrow {x^2} - {y^2} = - 3xy$$

So, combined equation of angle bisector is $${x^2} + 3xy - {y^2} = 0$$

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