JEE Main 2021 — Mathematics Coordinate Geometry

Given:Area of triangle formed by (5,6),(3,2)&(\alpha ,\beta )=12\mathrm{square}\mathrm{units}

$$\therefore \frac{1}{2}|\begin{matrix}\alpha & \beta & 1 \\ 5 & 6 & 1 \\ 3 & 2 & 1\end{matrix}|=12$$∴21​∣α53​β62​111​∣=12

$$\Rightarrow |\begin{matrix}\alpha & \beta & 1 \\ 5 & 6 & 1 \\ 3 & 2 & 1\end{matrix}|=24$$⇒∣α53​β62​111​∣=24

$$\Rightarrow \alpha (6-2)-\beta (5-3)+1(10-18)=\pm 24$$⇒α(6−2)−β(5−3)+1(10−18)=±24

$$\Rightarrow 4\alpha -2\beta -8=\pm 24$$⇒4α−2β−8=±24

$$\Rightarrow 4\alpha -2\beta =32,4\alpha -2\beta +16=0$$⇒4α−2β=32,4α−2β+16=0

$$\Rightarrow 2\alpha -\beta -16=0,2\alpha -\beta +8=0$$⇒2α−β−16=0,2α−β+8=0

So point will be either $(\alpha ,2\alpha +8),(\alpha ,2\alpha -16)$(α,2α+8),(α,2α−16)

For the point $(\alpha ,2\alpha +8)$(α,2α+8),

Distance from origin

$$D=\sqrt{{\alpha }^{2}+(2\alpha +8{)}^{2}}=\sqrt{5{\alpha }^{2}+32\alpha +64}(\because \beta =2\alpha +8)$$D=α2+(2α+8)2​=5α2+32α+64​(∵β=2α+8)

$$\Rightarrow {D}^{2}=5{\alpha }^{2}+32\alpha +64$$⇒D2=5α2+32α+64

For maximum or minimum length,

$$\frac{d({D}^{2})}{d\alpha }=10\alpha +32=0$$dαd(D2)​=10α+32=0

$$\Rightarrow \alpha =-\frac{16}{5}$$⇒α=−516​

$$\therefore \beta =\frac{-32}{5}+8=\frac{8}{5}$$∴β=5−32​+8=58​

$$\therefore D=\sqrt{{(-\frac{16}{5})}^{2}+{(\frac{8}{5})}^{2}}=\frac{8}{5}\sqrt{5}=\frac{8}{\sqrt{5}}$$∴D=(−516​)2+(58​)2​=58​5​=5​8​

Similarly if $\beta =2\alpha -16,D=\sqrt{{\alpha }^{2}+{(2\alpha -16)}^{2}}=\sqrt{5{\alpha }^{2}-64\alpha +256}$β=2α−16,D=α2+(2α−16)2​=5α2−64α+256​

$$\therefore {D}^{2}=5{\alpha }^{2}-64\alpha +256$$∴D2=5α2−64α+256

For maximum or minimum length,

$$\frac{d({D}^{2})}{d\alpha }=0$$dαd(D2)​=0

$$\Rightarrow 10\alpha -64=0$$⇒10α−64=0

$$\therefore \alpha =\frac{32}{5}$$∴α=532​

$\therefore D=\sqrt{{(\frac{32}{5})}^{2}+{(\frac{64}{5}-16)}^{2}}=\frac{16}{\sqrt{5}}$∴D=(532​)2+(564​−16)2​=5​16​at $\alpha =\frac{32}{5}$α=532​

So, least possible length of line segment is $\frac{8}{\sqrt{5}}$5​8​