JEE Main 2023 — Mathematics Coordinate Geometry

Given,

The triangles with vertices $A(2,1),B(0,0)$A(2,1),B(0,0) and $C(t,4),t=[0,4]$C(t,4),t=[0,4],

And the maximum and the minimum perimeters of such triangles are obtained at $t=\alpha$t=α and $t=\beta$t=β respectively,

Now to minimise $CA+CB$CA+CB, in below diagram take image of $B$B in $y=4$y=4,

We get, ${B}^{'}=(0,8)$B′=(0,8)

Now finding, equation of $A{B}^{'}$AB′ we get,

$$y-8=\frac{-7}{2}(x-0)$$y−8=2−7​(x−0)

Now, putting $y=4$y=4 in above equation we get,

$$-4=\frac{-7}{2}(x)$$−4=2−7​(x)

$$\Rightarrow x=\frac{8}{7}\Rightarrow \beta =\frac{8}{7}$$⇒x=78​⇒β=78​

Now, maximum perimeter will be possible if $\alpha =0$α=0 or $4$4

Now taking $\alpha =0$α=0 we get,

AB=\sqrt{5},BC=4&AC=\sqrt{13}

Now when $\alpha =4$α=4 we get,

AB=\sqrt{5},BC=4\sqrt{2}&AC=\sqrt{13}

Now on comparing the perimeter we get, maximum perimeter at $\alpha =4$α=4

Hence, $6\alpha +21\beta =48$6α+21β=48