The line joining the points $(-1,3)$ and $(4,-2)$ will pass through the point $(p,q)$ if
(a) $p-q=1$
(b) $p+q=1$
(c) $p-q=2$
(d) $p+q=2$

Hint: Write the general equation of the line in slope-intercept form which is $y=mx+c$, where $m$ is the slope of the line, that is, inclination of the line with respect to the x-axis, and $c$ is the intercept of the line on y-axis. Find the slope of the line using the formula $m=\dfrac{\Delta y}{\Delta x}$ and substitute this value in the equation to find $c$. Finally, substitute the value of coordinates $(p,q)$ in the equation of line to get the answer.

Complete Step-by-Step solution:
Let us assume that the equation of line is $y=mx+c$, where $m$ is the slope of the line, that is, inclination of the line with respect to the x-axis, and $c$ is the intercept of the line on y-axis.
Now, we know that slope of a line is given as:
$m=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}$. Here, ${{y}_{1}}=3,{{y}_{2}}=-2,{{x}_{1}}=-1\text{ and }{{x}_{2}}=4$. Therefore, substituting these values we get,
$m=\dfrac{3-(-2)}{-1-4}=\dfrac{5}{-5}=-1$
Therefore, the equation of line becomes, $y=-x+c$.
To determine the value of $c$, substitute the $(-1,3)$ in the equation of the line. Therefore,
$\begin{align}
  & 3=-(-1)+c \\
 & 1+c=3 \\
 & \Rightarrow c=2 \\
\end{align}$
Therefore the equation of line passing through $(-1,3)$ and $(4,-2)$ is $y=-x+2$.
Now, it is given that this line also passes through $(p,q)$, that means it will satisfy these coordinates. So, substituting these coordinates in the equation of line, we get,
$\begin{align}
  & q=-p+2 \\
 & \Rightarrow p+q=2 \\
\end{align}$
Hence, option (d) is the correct answer.

Note: One may note that, we have substituted the coordinates $(-1,3)$ in the equation of the line to determine the value of $c$. You may also substitute the other given coordinate $(4,-2)$ to determine the value of $c$. This will not change the result. So, the basic rule is, if any line is passing through a given point then the equation of the line must satisfy the point.