Question

write the equation of the line with slope $m = 0$ what passes through the point $(2,3)$
$\bigcirc y=2$
$\bigcirc y=3$
$\bigcirc x=2$
$\bigcirc x=3$
$\bigcirc y = 2x + 3$
$\bigcirc y = 3x + 2$
question 15 (4 points)
find the equation of the line that passes through the points $(4,6)$ and $(-2,-12)$
$\bigcirc y = -3x + 18$
$\bigcirc y + 6 = -3(x + 4)$
$\bigcirc y = 3x - 6$
$\bigcirc y + 4 = \frac{1}{3}(x + 2)$
$\bigcirc y = -3x - 18$

Explanation:

Step1: Identify horizontal line form

A line with slope $m=0$ is horizontal, with equation $y = k$ (where $k$ is a constant).

Step2: Substitute point (2,3)

Substitute $y=3$ into $y=k$, so $k=3$. Equation: $y=3$.
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Step1: Calculate slope of the line

Use slope formula $m=\frac{y_2-y_1}{x_2-x_1}$. Substitute $(4,6)$ and $(-2,-12)$:
$m=\frac{-12-6}{-2-4}=\frac{-18}{-6}=3$

Step2: Use point-slope form

Use point $(4,6)$: $y-6=3(x-4)$
Simplify to slope-intercept form:
$y-6=3x-12$
$y=3x-6$

Answer:

1. $y=3$
2. $y=3x-6$