We are asked to find the equation of a line in slope-intercept form, $y = mx + b$, given two points that the line passes through. We will solve for problems 12 and 13.

1. Problem Description

We are asked to find the equation of a line in slope-intercept form, $y = mx + b$, given two points that the line passes through. We will solve for problems 12 and

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2. Solution Steps

Problem 12:

Given points are $(-5, 3)$ and $(1, 3)$.

First, we find the slope, $m$:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{3 - 3}{1 - (-5)} = \frac{0}{6} = 0$

Since the slope is 0, the line is horizontal.

Now, we can use the slope-intercept form $y = mx + b$. Since the slope $m = 0$, the equation becomes $y = 0x + b$, or $y = b$.

Since both points have a $y$-coordinate of 3, the line is $y = 3$.

Therefore, $b = 3$.

The equation is $y = 0x + 3$, or $y = 3$.

Problem 13:

Given points are $(0, 1)$ and $(3, -2)$.

First, we find the slope, $m$:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{-2 - 1}{3 - 0} = \frac{-3}{3} = -1$

Now, we can use the slope-intercept form $y = mx + b$.

We have $m = -1$, so $y = -1x + b$.

We can use the point $(0, 1)$ to solve for $b$.

$1 = -1(0) + b$

$1 = 0 + b$

$b = 1$

The equation is $y = -x + 1$.

3. Final Answer

Problem 12: $y = 3$

Problem 13: $y = -x + 1$