Finding the Equation of a Line: Practice

Point-slope form is built from a point $(x_1, y_1)$ and the slope $m$, subtracting on both the $y$-side and the $x$-side.

$y - y_1 = m(x - x_1)$

The choice $m = \tfrac{y_2 - y_1}{x_2 - x_1}$ is the slope formula, not the equation of the line.

Substitute the point $(2, 3)$ and slope $4$ into $y - y_1 = m(x - x_1)$.

$y - 3 = 4(x - 2)$

Both $y_1 = 3$ and $x_1 = 2$ are subtracted.

In point-slope form $y - y_1 = m(x - x_1)$, the point is $(x_1, y_1)$, read straight from the subtractions.

$y - 4 = 3(x - 2) \;\Rightarrow\; (x_1, y_1) = (2, 4)$

So the line passes through $(2, 4)$.

A horizontal line keeps the same height at every point, so its equation fixes $y$ at that height. Through $(3, 6)$ the height is $6$.

$y = 6$

The equation $x = 3$ would be the vertical line through the point.

A vertical line keeps the same $x$-coordinate at every point, so its equation fixes $x$. Through $(3, 6)$ the $x$-value is $3$.

$x = 3$

The equation $y = 6$ would be the horizontal line through the point.

Point-slope form $y - y_1 = m(x - x_1)$ is built from exactly two ingredients: one point $(x_1, y_1)$ and the slope $m$.

$\text{point } (x_1, y_1) \;+\; \text{slope } m \;\Rightarrow\; y - y_1 = m(x - x_1)$

A single point together with the slope is enough to pin the line down.

Match to $y - y_1 = m(x - x_1)$. The term $y + 3$ is $y - (-3)$, so $y_1 = -3$, and $x - 5$ gives $x_1 = 5$.

$y + 3 = 2(x - 5) \;\Rightarrow\; (x_1, y_1) = (5, -3)$

So the line passes through $(5, -3)$.

An equation of the form $y = c$ fixes the height at $c$ for every $x$, which is a horizontal line.

$y = -2 \;\Rightarrow\; \text{height } -2 \text{ at every point}$

So the line is horizontal, two units below the $x$-axis.

Substitute the point $(0, 5)$ and slope $2$ into point-slope form, where $x_1 = 0$.

$y - 5 = 2(x - 0) = 2x$

So the equation is $y - 5 = 2x$.

Test each point in the equation. For $(3, 4)$,

$4 - 1 = 3(3 - 2) \;\Rightarrow\; 3 = 3 \;\checkmark$

So $(3, 4)$ is on the line, while the other points leave the sides unequal.

A horizontal line fixes the height. Through $(-4, 3)$ the height is $3$.

$y = 3$

The equation $x = -4$ would be the vertical line through the point.

Substitute the point $(3, 4)$ and slope $2$.

$y - 4 = 2(x - 3)$

Here $x_1 = 3$ and $y_1 = 4$.