Finding the Equation of a Line: Additional Practice (Set 1)

Find the slope, then use point-slope form with $(-1, 5)$.

$m = \frac{-3 - 5}{3 - (-1)} = \frac{-8}{4} = -2$

With slope $-2$ and $(-1, 5)$: $y - 5 = -2(x + 1)$. Each distractor changes a sign and gives a different line.

Start from point-slope form and clear the fraction.

$y - 3 = \tfrac{3}{4}(x - 2), \qquad 4(y - 3) = 3(x - 2), \qquad 4y - 12 = 3x - 6$

Collecting terms gives $3x - 4y = -6$. Check $(2, 3)$: $3(2) - 4(3) = 6 - 12 = -6$.

The line passes through $(0, 3)$ on the $y$-axis and $(4, 0)$ on the $x$-axis. Its slope is

$m = \frac{0 - 3}{4 - 0} = -\tfrac{3}{4}, \qquad y - 3 = -\tfrac{3}{4}(x - 0),$

and clearing the fraction gives $3x + 4y = 12$. Check both marked points: $3(0) + 4(3) = 12$ and $3(4) + 4(0) = 12$.

Write point-slope form, then substitute $x = 1$.

$y - 2 = 3(x + 1), \qquad k - 2 = 3(1 + 1) = 6, \qquad k = 8$

So $k = 8$.

Find the slope, write point-slope form, then convert.

$m = \frac{3 - (-3)}{5 - 2} = \frac{6}{3} = 2, \qquad y + 3 = 2(x - 2)$

Collecting terms gives $2x - y = 7$. Check $(5, 3)$: $2(5) - 3 = 7$.

Start from point-slope form and clear the fraction.

$y + 1 = -\tfrac{1}{3}(x - 6), \qquad 3(y + 1) = -(x - 6), \qquad 3y + 3 = -x + 6$

Collecting terms gives $x + 3y = 3$. Check $(6, -1)$: $6 + 3(-1) = 3$.

Match to $y - y_1 = m(x - x_1)$. The term $y - 7$ gives $y_1 = 7$; the term $x + 9$ is $x - (-9)$, so $x_1 = -9$; and $m = \tfrac{2}{3}$.

$y - 7 = \tfrac{2}{3}(x + 9) \;\Rightarrow\; (x_1, y_1) = (-9, 7), \; m = \tfrac{2}{3}$

So the point is $(-9, 7)$ with slope $\tfrac{2}{3}$.

A vertical line fixes the $x$-coordinate. Through $(-4, 6)$ the $x$-value is $-4$.

$x = -4$

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

Find the slope, write point-slope form, and clear the fraction.

$m = \frac{4 - 1}{4 - (-2)} = \frac{3}{6} = \tfrac{1}{2}, \qquad 2(y - 1) = x + 2, \qquad 2y - 2 = x + 2$

Collecting terms gives $x - 2y = -4$. Check $(4, 4)$: $4 - 2(4) = -4$.

Point-slope form with slope $-2$ and the point $(3, 4)$ gives

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

The distractors change the sign of a coordinate or the slope, so each is a different line.

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

$3(2) - 2k = 4, \qquad 6 - 2k = 4, \qquad -2k = -2, \qquad k = 1$

So $k = 1$.

The line passes through $(0, -2)$ on the $y$-axis and $(3, 0)$ on the $x$-axis. Its slope is

$m = \frac{0 - (-2)}{3 - 0} = \tfrac{2}{3}, \qquad y + 2 = \tfrac{2}{3}(x - 0),$

and clearing the fraction gives $2x - 3y = 6$. Check both marked points: $2(0) - 3(-2) = 6$ and $2(3) - 3(0) = 6$.