1. **Problem 1: Find the equation of the straight line EF passing through points E(-1, 3) and F(6, -1).** 2. The formula for the slope (gradient) $m$ of a line through points $(x_1, y_1)$ and $(x_2, y_2)$ is: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 3. Calculate the slope: $$m = \frac{-1 - 3}{6 - (-1)} = \frac{-4}{7} = -\frac{4}{7}$$ 4. Use the point-slope form of the line equation: $$y - y_1 = m(x - x_1)$$ Using point E(-1, 3): $$y - 3 = -\frac{4}{7}(x + 1)$$ 5. Simplify: $$y - 3 = -\frac{4}{7}x - \frac{4}{7}$$ $$y = -\frac{4}{7}x - \frac{4}{7} + 3 = -\frac{4}{7}x + \frac{17}{7}$$ 6. Multiply both sides by 7 to clear the denominator: $$7y = -4x + 17$$ 7. The correct equation is option A: $7y = -4x + 17$. --- 8. **Problem 2: Given point J(-2, 3) and gradient of line JK is 2, find possible coordinates of point K.** 9. The slope formula is: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 10. Let point K be $(x, y)$. Using $m=2$ and point J(-2, 3): $$2 = \frac{y - 3}{x - (-2)} = \frac{y - 3}{x + 2}$$ 11. Rearranged: $$y - 3 = 2(x + 2)$$ $$y = 2x + 4 + 3 = 2x + 7$$ 12. Check each option: - A: (-1, 1): $y = 1$, $2x + 7 = 2(-1) + 7 = 5$ (No) - B: (-1, 3): $y = 3$, $2(-1) + 7 = 5$ (No) - C: (-3, 1): $y = 1$, $2(-3) + 7 = 1$ (Yes) - D: (-3, 2): $y = 2$, $2(-3) + 7 = 1$ (No) 13. The correct answer is C: (-3, 1). --- 14. **Problem 3: Find factors of $2p^2 + 7p + 3$.** 15. To factor quadratic $ax^2 + bx + c$, find two numbers that multiply to $a \times c = 2 \times 3 = 6$ and add to $b = 7$. 16. Numbers are 6 and 1. 17. Rewrite middle term: $$2p^2 + 6p + p + 3$$ 18. Factor by grouping: $$2p(p + 3) + 1(p + 3) = (2p + 1)(p + 3)$$ 19. The correct factorization is option D: $(2p + 1)(p + 3)$. --- **Final answers:** - Problem 1: A - Problem 2: C - Problem 3: D