1. **Problem 1:** For each pair of points, find the length of the line segment connecting them. Recall the distance formula between two points $ (x_1, y_1) $ and $ (x_2, y_2) $: $$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ (a) Points: (1,7) and (5,7) $$\sqrt{(5-1)^2 + (7-7)^2} = \sqrt{4^2 + 0} = \sqrt{16} = 4$$ (b) Points: (8,5) and (-1,7) $$\sqrt{(-1-8)^2 + (7-5)^2} = \sqrt{(-9)^2 + 2^2} = \sqrt{81 + 4} = \sqrt{85} \approx 9.22$$ (c) Points: (-2,2) and (4,3) $$\sqrt{(4+2)^2 + (3-2)^2} = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37} \approx 6.08$$ (d) Points: (-1,6) and (-3,-2) $$\sqrt{(-3+1)^2 + (-2-6)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68} \approx 8.25$$ (e) Points: (4,-1) and (9,7) $$\sqrt{(9-4)^2 + (7+1)^2} = \sqrt{5^2 + 8^2} = \sqrt{25 + 64} = \sqrt{89} \approx 9.43$$ (f) Points: (-1,7) and (4,7) Same as (a): distance = 5 (g) Points: (4,8) and (4,3) $$\sqrt{(4-4)^2 + (3-8)^2} = \sqrt{0 + (-5)^2} = 5$$ (h) Points: (-4,-3) and (5,6) $$\sqrt{(5+4)^2 + (6+3)^2} = \sqrt{9^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} \approx 12.73$$ (i) Points: (-9,-6) and (-8,-2) $$\sqrt{(-8+9)^2 + (-2+6)^2} = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.12$$ --- 2. **Problem 2:** Given the triangle with points A(1,7), B(-3,4), C(6,3): (a) Find length of each side: - AB: $$\sqrt{(-3-1)^2 + (4-7)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ - BC: $$\sqrt{(6+3)^2 + (3-4)^2} = \sqrt{9^2 + (-1)^2} = \sqrt{81 + 1} = \sqrt{82} \approx 9.06$$ - AC: $$\sqrt{(6-1)^2 + (3-7)^2} = \sqrt{5^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \approx 6.40$$ (b) Check if triangle ABC is isosceles: An isosceles triangle has at least two equal sides. Here, no two sides are equal (5, 9.06, 6.40), so it is NOT isosceles. (c) Check if triangle is right angled at A: Use Pythagoras theorem: check if $AB^2 + AC^2 = BC^2$ $$5^2 + 6.4^2 = 25 + 40.96 = 65.96$$ $$BC^2 = 9.06^2 = 82.03$$ Since $65.96 \neq 82.03$, triangle is NOT right angled at A. Area formula if right angled at A would be: $$\frac{1}{2} \times AB \times AC$$ but since not right angled here, area calculation needs Heron's formula (not requested). --- 3. **Problem 3:** Plot points and discuss properties (not fully detailed here as plotting is requested but no graph plotting is included). The points mentioned relate to properties of rhombus, parallelogram, and right angled triangle based on given vertices positions. --- 4. **Problem 4:** Find unknown values relating to distances and points. Details incomplete, so cannot solve explicitly. --- 5. **Problem 5:** Find equations of lines. (a) Line with slope $m = -3$ passing through $(-1, -2)$: Use point-slope form: $$y - y_1 = m(x - x_1)$$ $$y + 2 = -3(x + 1)$$ $$y + 2 = -3x - 3$$ $$y = -3x - 5$$ (b) Line cutting x-axis at 4 and y-axis at 5: Intercept form: $$\frac{x}{4} + \frac{y}{5} = 1$$ Multiply both sides by 20: $$5x + 4y = 20$$ or solve for y: $$y = 5 - \frac{5}{4}x$$ (c) Line cutting x-axis at 2 and passes through (-5): Assuming point (-5, y) is (-5, 0) or missing y-coordinate, insufficient data to solve. --- 6. **Problem 8:** Similar to problem 5 but with slope 3 and points given. (a) Slope 3 passing through (-1,-2): $$y + 2 = 3(x + 1)$$ $$y = 3x + 3 - 2$$ $$y = 3x + 1$$ (b) Line cuts x-axis at 4 and y-axis at 3: $$\frac{x}{4} + \frac{y}{3} = 1$$ $$3x + 4y = 12$$ (c) Line cuts x-axis at -2 and passes through (2, -5): Intercept form: $$y = m x + c$$ At x-intercept (-2,0): $$0 = m (-2) + c \Rightarrow c = 2m$$ Using point (2,-5): $$-5 = m (2) + c = 2m + 2m = 4m$$ $$4m = -5 \Rightarrow m = -\frac{5}{4}$$ So: $$c = 2 (-\frac{5}{4}) = -\frac{5}{2}$$ Equation: $$y = -\frac{5}{4} x - \frac{5}{2}$$