Answer
$y=\displaystyle \frac{s-q}{r-p}x+\frac{qr-ps}{r-p}$
Work Step by Step
Target: $y=mx+b,$ (find m and b).
Given $(x_{1},y_{1})=(p,q)$ and $(x_{2},y_{2})=(r,s)$ on a line,
calculate the slope :
$m=\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$=\displaystyle \frac{s-q}{r-p} \qquad...(r\neq p)$
So, the form of our linear equation is $y=\displaystyle \frac{s-q}{r-p}x+b$
To find b, substitute the coordinates of one of the given points and solve for b.
$q=\displaystyle \frac{s-q}{r-p}(p)+b\qquad /-\frac{p(s-q)}{r-p}$
$q-\displaystyle \frac{p(s-q)}{r-p}=b$
$b=\displaystyle \frac{q(r-p)-p(s-q)}{r-p}=\frac{qr-qp-ps+pq}{r-p}=\frac{qr-ps}{r-p}$
Thus,
$y=\displaystyle \frac{s-q}{r-p}x+\frac{qr-ps}{r-p}$
