Question: User: Topic 2 Test Review 1. What is the equation of the line shown in point-slope form? A. y - 1 = 2(x + 2) B. y - 1 = -2(x + 2) C. y + 1 = -2(x - 2) D. y + 1 = 2(x - 2) 2. Write the equation in slope-intercept form of a line parallel to y = -6x + 5 that passes through (-1, -1) y = x + graph: A coordinate grid in the top-right shows a line slanting downward from left to right, crossing the y-axis at about -4 and passing through points near (-2, 0) and (1, -6); position_hint=top-rightexplain number one this step by step

1. **Problem:** Find the equation of the line shown in point-slope form from the given options. 2. **Recall the point-slope form formula:** $$y - y_1 = m(x - x_1)$$ where $m$ is the slope and $(x_1, y_1)$ is a point on the line. 3. **Identify the slope $m$ and point $(x_1, y_1)$ from the graph:** - The line passes through points approximately $(-2, 0)$ and $(1, -6)$. - Calculate the slope: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - 0}{1 - (-2)} = \frac{-6}{3} = -2$$ 4. **Choose a point on the line to use in the formula:** - Let's use the point $(-2, 0)$. 5. **Plug values into point-slope form:** $$y - 0 = -2(x - (-2))$$ which simplifies to $$y - 0 = -2(x + 2)$$ or $$y = -2(x + 2)$$ 6. **Rewrite to match the options:** - Subtract $1$ from both sides to compare with options that have $y - 1$ or $y + 1$: Since the point used is $(-2, 0)$, options with $y - 1$ or $y + 1$ correspond to points $(x_1, y_1)$ with $y_1 = 1$ or $-1$. - Check option B: $y - 1 = -2(x + 2)$ - This corresponds to point $(-2, 1)$, slope $-2$. - The graph shows the line passes near $(-2, 0)$, so option B is close but point is off by 1. - Check option C: $y + 1 = -2(x - 2)$ - This corresponds to point $(2, -1)$, slope $-2$. - The graph line passes through $(-2, 0)$ and $(1, -6)$, so this point is not on the line. - Check option D: $y + 1 = 2(x - 2)$ - Slope is positive 2, but our slope is $-2$, so discard. - Check option A: $y - 1 = 2(x + 2)$ - Slope is positive 2, discard. 7. **Conclusion:** The correct equation in point-slope form matching the slope and a point on the line is: $$y - 0 = -2(x + 2)$$ which is not exactly listed but closest to option B if the point was $( -2, 1)$ instead of $( -2, 0)$. Since the graph shows the line crossing near $(-2, 0)$, option B is the best match for slope and form. **Final answer:** B. $y - 1 = -2(x + 2)$ --- **Note:** The question asks to explain number one step by step, so only problem 1 is solved here.