What to find the path difference in constructive interference? 
Hi there,
I am unsure of how to calculate the path difference in questions relating to constructive and destructive interference.
I understand that when two waves are exactly in phase with each other, constructive interference occurs, so I labeled the intersecting points as areas where constructive interference occurs. But the answer scheme states that at the points where the waves are in phase, the path difference is $nλ$, while at the points where destructive interference occurs, the path difference is $(n − 1/2)\lambda$.
Where do these equations come from?
 A: Re. 
"Where do these equations come from?"
Note that $sin(\theta + n\lambda)$ = $sin(\theta)$ so a phase shift of $n\lambda$ leaves the two waves in phase so they reinforce.
Whereas $sin(\theta + (n-1/2)\lambda)$ = $-sin(\theta)$ so a phase shift of $(n-1/2)\lambda$ leaves the two waves 180 degrees out of phase so the cancel.
A: You probably know that the total wave in space is simply the sum of the waves from points $A$ and $B$. What matters for the strength of a signal is the intensity of a wave, as its squared modulus. Then, the maximum of constructive interference (point X) occurs when two peaks - represented by red circles - or two troughs - represented by green circles - meet. On the other hand, the minimum (point Y) will be where a peak meets a trough.
As you can see here, in a somewhat different problem: at $x=0$, both waves have values $w_1(0)=w_2(0)=1$, so they interfere constructively, while at $x=\pi$ we have  $w_1(\pi)=-w_2(\pi)=-1$, so the intensity is 0 and the interference is destructive.

The other part of your question is about the path differences. In your specific case, notice that the peaks happen at integer multiples $n\lambda$ of the wavelength, and the troughs happen at semi-integers $(n-1/2)\lambda$ away from the source. If a maximum is at the intersection of the red circles, the path difference is 
$$
d = n_A\lambda - n_B\lambda = (n_A-n_B)\lambda=n\lambda,\ \text{ with }n\text{ integer},
$$
and if it is at a intersection of green circles,
$$
d = \left(n_A-\frac{1}{2}\right)\lambda - \left(n_B-\frac{1}{2}\right)\lambda = \left(n_A-n_B\right)\lambda=n\lambda,\ \text{ with }n\text{ integer}.
$$
If it is at a minimum, it is at a point where a red circle meets a green circle, so we have a situation like
$$
d = \left(n_A-\frac{1}{2}\right)\lambda -n_B\lambda = \left(n_A-n_B-\frac{1}{2}\right)\lambda=\left(n-\frac{1}{2}\right)\lambda,\ \text{ with }n\text{ integer}.
$$
Notice that the $-1/2$ might be on the other parentheses, it doesn't change the result.
A: Let say you have two electromagnetic wave
$$y_{1}=A\sin(\omega t-kx_{1}),\,y_{2}=A\sin(\omega t-kx_{2}),$$
Now phase difference denoted, by variable $h$, so we get 
$$kx_{1}-kx_{2}=h\tag{1}$$
Where $k$ denote wave number, because we measure the distance traveled by wave in terms of wavelength, so we get the wave number. 
Where $(x_{1}-x_{2})$ is the path difference.
Two waves interfere like a vector did. 
So when the phase difference is integral of $2n(\pi)$,then wave will show constructive interference, if it $(n+1/2)(\pi)$ then they show destructive interference, after this value in equation 1, you will get path difference between waves. As a function of $n$. 
A: They come from the realisation that a sine or cosine function changes sign twice every wavelength.  
