1
Numerical

### JEE Main 2021 (Online) 1st September Evening Shift

A man starts walking from the point P($-$3, 4), touches the x-axis at R, and then turns to reach at the point Q(0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $5\left( {{{(PR)}^2} + {{(RQ)}^2}} \right)$ is equal to ____________.

## Explanation 50(PR2 + RQ2)

= 50(20 + 5)

= 50(25)

= 1250
2
Numerical

### JEE Main 2021 (Online) 1st September Evening Shift

Let the points of intersections of the lines x $-$ y + 1 = 0, x $-$ 2y + 3 = 0 and 2x $-$ 5y + 11 = 0 are the mid points of the sides of a triangle ABC is _____________.

## Explanation

Intersection point of given lines are (1, 2), (7, 5), (2, 3) $\Delta = {1 \over 2}\left| {\matrix{ 1 & 2 & 1 \cr 7 & 5 & 1 \cr 2 & 3 & 1 \cr } } \right|$

$= {1 \over 2}[1(5 - 3) - 2(7 - 2) + 1(21 - 10)]$

$= {1 \over 2}[2 - 10 + 11]$

$\Delta$DEF $= {1 \over 2}(3) = {3 \over 2}$

$\Delta$ABC = 4$\Delta$DEF $= 4\left( {{3 \over 2}} \right) = 6$
3
Numerical

### JEE Main 2021 (Online) 20th July Evening Shift

Consider a triangle having vertices A($-$2, 3), B(1, 9) and C(3, 8). If a line L passing through the circum-centre of triangle ABC, bisects line BC, and intersects y-axis at point $\left( {0,{\alpha \over 2}} \right)$, then the value of real number $\alpha$ is ________________.

## Explanation ${\left( {\sqrt {50} } \right)^2} = {\left( {\sqrt {45} } \right)^2} + {\left( {\sqrt 5 } \right)^2}$

$\angle B = 90^\circ$

Circum-center $= \left( {{1 \over 2},{{11} \over 2}} \right)$

Mid point of BC $= \left( {2,{{17} \over 2}} \right)$

Line : $\left( {y - {{11} \over 2}} \right) = 2\left( {x - {1 \over 2}} \right) \Rightarrow y = 2x + {9 \over 2}$

Passing through $\left( {0,{\alpha \over 2}} \right)$

${\alpha \over 2} = {9 \over 2} \Rightarrow \alpha = 9$
4
Numerical

### JEE Main 2021 (Online) 18th March Morning Shift

A square ABCD has all its vertices on the curve x2y2 = 1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is

## Explanation

x2y2 = 1

$\Rightarrow$ y2 = ${1 \over {{x^2}}}$

$\Rightarrow$ y = $\pm {1 \over x}$

Graph of this equation, $OA \bot OB$

$\Rightarrow \left( {{1 \over {{p^2}}}} \right)\left( { - {1 \over {{q^2}}}} \right) = - 1$

$\Rightarrow {p^2}{q^2} = 1$

$P\left( {{{p + q} \over 2},{{{1 \over p} - {1 \over q}} \over 2}} \right)$ midpoint of AB lies

On ${x^2}{y^2} = 1$

$\Rightarrow {(p + q)^2}{\left( {{1 \over p} - {1 \over q}} \right)^2} = 16$

$\Rightarrow {(p + q)^2}{(p - q)^2} = 16$

$\Rightarrow {({p^2} - {q^2})^2} = 16$

$\Rightarrow {P^2} - {1 \over {{P^2}}} = \pm 4$

$\Rightarrow {p^4} \pm 4{p^2} - 1 = 0$

$\Rightarrow {p^2} = {{ \pm 4 \pm \sqrt {20} } \over 2} = \pm 2 \pm \sqrt 5$

$\Rightarrow {p^2} = 2 + \sqrt 5$ or $- 2 + \sqrt 5$

$O{B^2} = {p^2} + {1 \over {{p^2}}} = 2 + \sqrt 5 + {1 \over {2 + \sqrt 5 }}$ or $- 2 + \sqrt 5 + {1 \over { - 2 + \sqrt 5 }} = 2\sqrt 5$

Area $= 4\left( {{1 \over 2}} \right)(OA)(OB) = 2{(OB)^2} = 4\sqrt 5$