Finding the Equation of a Line: Practice

Substitute each point and keep the one that makes $3x + y = 7$ true. For $(2, 1)$,

$3(2) + 1 = 6 + 1 = 7 \;\checkmark$

So $(2, 1)$ is on the line; the others give $6$, $3$, and $10$.

Match to $y - y_1 = m(x - x_1)$. The term $y + 6$ is $y - (-6)$, so $y_1 = -6$, and $x - 2$ gives $x_1 = 2$.

$y + 6 = -4(x - 2) \;\Rightarrow\; (x_1, y_1) = (2, -6)$

So the line passes through $(2, -6)$.

A horizontal line fixes the height at every point. Through $(-1, -5)$ the height is $-5$.

$y = -5$

The equation $x = -1$ would be the vertical line through the point.

Clear the fraction by multiplying both sides by $3$, then distribute and collect terms.

$3(y - 2) = x - 1, \qquad 3y - 6 = x - 1, \qquad x - 3y = -5$

So standard form is $x - 3y = -5$, with $A = 1$.

The two points share the same height $y = 5$, so the line is horizontal.

$m = \frac{5 - 5}{6 - 2} = \frac{0}{4} = 0 \;\Rightarrow\; y = 5$

A slope of $0$ means the equation is simply $y = 5$.

The two points share the same $x$-coordinate $3$, so the run is $0$ and the slope is undefined: the line is vertical.

$x = 3$

A vertical line's equation just fixes $x$ at the shared value.

Distribute the slope, then move the $x$-term over.

$y + 2 = -3x + 3, \qquad 3x + y + 2 = 3, \qquad 3x + y = 1$

So standard form is $3x + y = 1$.

Find the slope, then use point-slope form with the point $(5, 2)$.

$m = \frac{-6 - 2}{1 - 5} = \frac{-8}{-4} = 2$

With slope $2$ and $(5, 2)$: $y - 2 = 2(x - 5)$.

Test each point in the equation. For $(2, 1)$,

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

So $(2, 1)$ lies on the line, while the others leave the sides unequal.

Standard form $Ax + By = C$ puts both variable terms on one side and the constant on the other, with integer coefficients.

$3x + 2y = 6$

The others are in point-slope form, solved for $y$, or have a variable on the wrong side.

Substitute the point $(-2, -3)$ and slope $-1$. Both coordinates are negative, so $y - (-3) = y + 3$ and $x - (-2) = x + 2$.

$y + 3 = -1(x + 2) = -(x + 2)$

So the equation is $y + 3 = -(x + 2)$.

Substitute $y = 2$ and solve for $x = k$.

$k + 2(2) = 10, \qquad k + 4 = 10, \qquad k = 6$

So $k = 6$.