<s>[INST] Three vertices of a triangle are (2,7),(4,-1),(0,Y).if the perimeter of the triangle is the least then express Y in the lowest rational form as (p/q). I don't know how to proceed in this question. What I know is that the perimeter is the sum of all the sides.
 [/INST] Notice, the length of side with vertices $(2, 7)$ & $(4, -1)$, using distance formula, is $$=\sqrt{(2-4)^2+(7+1)^2}=\sqrt{68}$$
Similarly, the length of side with vertices $(2, 7)$ & $(0, y)$ is $$=\sqrt{(2-0)^2+(7-y)^2}=\sqrt{(y-7)^2+4}$$ the length of side with vertices $(4, -1)$ & $(0, y)$ is $$=\sqrt{(4-0)^2+(-1-y)^2}=\sqrt{(y+1)^2+16}$$
Hence, the perimeter $P$ is the sum of all three sides hence. we have $$P=\sqrt{68}+\sqrt{(y-7)^2+4}+\sqrt{(y+1)^2+16}$$
Differentiating w.r.t. $y$, we get $$\frac{dP}{dy}=\frac{y-7}{\sqrt{(y-7)^2+4}}+\frac{y+1}{\sqrt{(y+1)^2+16}}$$
$$\frac{d^2P}{dy^2}=\frac{4}{((y-7)^2+4)^{3/2}}+\frac{16}{((y-7)^2+4)^{3/2}}$$
It is clear that $\frac{d^2P}{dy^2}>0\ \ \forall\ \ y\in R$ Hence, the perimeter is minimum/least
Now, setting $\frac{dP}{dy}=0$ for minimum, we get $$\frac{y-7}{\sqrt{(y-7)^2+4}}+\frac{y+1}{\sqrt{(y+1)^2+16}}=0$$
$$(y+1)^2=4(y-7)^2$$ $$y+1=\pm 2(y-7)$$
$$\implies y=15, \ \ \frac{13}{3} $$
But, it is clear that at both the values $y=15$ & $y=\frac{13}{5}$,  the perimeter $P$ of triangle is minimum/least. 

But at $y=15$ the triangle does not exist hence $y=15$ is unacceptable. 

Hence, we get  $$\color{red}{y=\frac{13}{3}} $$
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