Straight Line Definition

Question

Definition: Let $P(x_0, y_0)$ be a point and let $m$ be a real number. The line through $P$ with slope $m$ is the set of all points $Q(x, y)$ with,

$y -y_0 = m(x - x_0)$

Does the set of all points $Q(x, y)$ also includes the point $P(x_0, y_0)$ since its on the line?

Answer 1

Yes, indeed, Samama. Since we are creating the equation of the line using point P, we are assured that $P = (x_0, y_0)$ is on the line. Just substitute the coordinates of $P$ into the equation, and you will satisfy the equality given by the equation of the line. So $P$ is one of the points included on the line $L$ which is the set of all points determined by a given $m$, $P = (x_0, y_0)$: $$L = Q(x, y) = \{(x, y)\mid y -y_0 = m(x - x_0),\}$$

since $y_0 - y_0 = m(x_0 - x_0) \iff 0 = 0 \;\;\large \checkmark$

But $P$ is only one point in the infinite set defined by $Q(x, y)$. So don't mistaken $Q(x, y)$ as one point $Q$, it defines all points on the line determined by the given slope $m$, and a given point $P = (x_0, y_0)$ lying on the line.

Answer 2

Just set $x=x_0$, $y=x_0$. Your definition of line yields that $(x_0,y_0)$ lies on the line if and only if $$ y_0 - y_0 = m(x_0 - x_0) \text{.} $$ Which, of course, is true no matter the value of $x_0$, $y_0$ and $m$.