Write the equation of the line in slope-intercept form with the given point and slope or two points. \begin{tabular}{ll|l} 14. \( (-7,13) \); slope \( =-2 \) & 15. \( (-4,6) \); slope \( =-\frac{3}{4} \) & 14.\end{tabular} \( \begin{array}{lll}\text { 16. }(-5,-11) \text { and }(-2,1) & \text { 17. }(-6,8) \text { and }(3,-7) & 17 .\end{array} \)

We’re given four problems. Follow the steps below for each. ───────────────────────────── Problem 14. Given: point (–7, 13) and slope m = –2. 1. Use the point-slope form: y – y₁ = m(x – x₁). 2. Substitute the values: y – 13 = –2(x – (–7)) = –2(x + 7). 3. Distribute: y – 13 = –2x – 14. 4. Solve for y: y = –2x – 14 + 13 ⟹ y = –2x – 1. Answer: y = –2x – 1 ───────────────────────────── Problem 15. Given: point (–4, 6) and slope m = –¾. 1. Start with point-slope form: y – y₁ = m(x – x₁). 2. Substitute the given values: y – 6 = –¾(x – (–4)) = –¾(x + 4). 3. Distribute the slope: y – 6 = –¾x – (¾ × 4) = –¾x – 3. 4. Solve for y: y = –¾x – 3 + 6 ⟹ y = –¾x + 3. Answer: y = –(3/4)x + 3 ───────────────────────────── Problem 16. Given: points (–5, –11) and (–2, 1). 1. Find the slope m using: m = (y₂ – y₁)/(x₂ – x₁) = (1 – (–11))/(–2 – (–5)) = (12)/(3) = 4. 2. Use the point-slope form with one of the points (we’ll use (–5, –11)): y – (–11) = 4(x – (–5)), which simplifies to y + 11 = 4(x + 5). 3. Expand: y + 11 = 4x + 20. 4. Solve for y: y = 4x + 20 – 11 ⟹ y = 4x + 9. Answer: y = 4x + 9 ───────────────────────────── Problem 17. Given: points (–6, 8) and (3, –7). 1. Calculate the slope m: m = (–7 – 8)/(3 – (–6)) = (–15)/(9) = –5/3. 2. Use the point-slope form with point (–6, 8): y – 8 = –(5/3)(x – (–6)) = –(5/3)(x + 6). 3. Expand: y – 8 = –(5/3)x – (5/3 × 6). Note that (5/3 × 6) = 10. 4. So, y – 8 = –(5/3)x – 10. 5. Solve for y: y = –(5/3)x – 10 + 8 ⟹ y = –(5/3)x – 2. Answer: y = –(5/3)x – 2 ───────────────────────────── Final Answers: 14. y = –2x – 1 15. y = –(3/4)x + 3 16. y = 4x + 9 17. y = –(5/3)x – 2