1. **Verify that (2, 5) lies on the line with equation** $y = 3x - 1$. Step 1: Substitute $x=2$ into the equation: $$y = 3(2) - 1 = 6 - 1 = 5$$ Step 2: Since $y=5$ matches the point's $y$-coordinate, the point $(2,5)$ lies on the line. 2. **Which points lie on the line** $3x + 2y = 6$? Check each point by substituting $x$ and $y$: (a) $(3,0)$: $3(3) + 2(0) = 9 + 0 = 9 \neq 6$ (No) (b) $(-2,6)$: $3(-2) + 2(6) = -6 + 12 = 6$ (Yes) (c) $(2,0)$: $3(2) + 2(0) = 6 + 0 = 6$ (Yes) (d) $(8,0)$: $3(8) + 2(0) = 24 + 0 = 24 \neq 6$ (No) 3. **Find coordinates where lines intersect axes:** (i) For $y = 2x - 6$: - Intersection with x-axis: set $y=0$ $$0 = 2x - 6 \Rightarrow 2x = 6 \Rightarrow x = 3$$ Point: $(3,0)$ - Intersection with y-axis: set $x=0$ $$y = 2(0) - 6 = -6$$ Point: $(0,-6)$ (ii) For $2x + 3y = -6$: - Intersection with x-axis: set $y=0$ $$2x = -6 \Rightarrow x = -3$$ Point: $(-3,0)$ - Intersection with y-axis: set $x=0$ $$3y = -6 \Rightarrow y = -2$$ Point: $(0,-2)$ 4. **Find $k$ if point $(3,k)$ lies on line $2x + 3y - 6 = 0$**: Substitute $x=3$: $$2(3) + 3k - 6 = 0 \Rightarrow 6 + 3k - 6 = 0 \Rightarrow 3k = 0 \Rightarrow k = 0$$ 5. **Point A(2,6) and origin O(0,0):** (a) i. Gradient of line OA: $$m = \frac{6 - 0}{2 - 0} = \frac{6}{2} = 3$$ ii. Y-intercept of OA is the $y$-coordinate when $x=0$, which is 0 (origin). iii. Equation of line OA using $y = mx + c$: $$y = 3x + 0 = 3x$$ (b) Gradient of line: - Perpendicular to OA: $m_{\perp} = -\frac{1}{3}$ - Parallel to OA: $m_{\parallel} = 3$ 6. **Coordinates X(-8,-5), Y(-2,-1), M midpoint of XY:** (a) i. Gradient of XY: $$m = \frac{-1 - (-5)}{-2 - (-8)} = \frac{4}{6} = \frac{2}{3}$$ ii. Midpoint M: $$M = \left(\frac{-8 + (-2)}{2}, \frac{-5 + (-1)}{2}\right) = (-5, -3)$$ (b) Equation of perpendicular bisector: - Gradient perpendicular to XY: $$m_{\perp} = -\frac{3}{2}$$ - Using point-slope form with midpoint M(-5,-3): $$y - (-3) = -\frac{3}{2}(x - (-5))$$ $$y + 3 = -\frac{3}{2}(x + 5)$$ 7. **Graph AB passes through A(0,2) and B(3,3):** (a) Value of $y$ when: - $x=0$: $y=2$ (point A) - $x=3$: $y=3$ (point B) (b) (i) Complete table for $2y + 3x = 4$: For $x=-1$: $$2y + 3(-1) = 4 \Rightarrow 2y - 3 = 4 \Rightarrow 2y = 7 \Rightarrow y = \frac{7}{2} = 3.5$$ For $x=2$: Given $y = -1$ (already filled) For $x=4$: $$2y + 3(4) = 4 \Rightarrow 2y + 12 = 4 \Rightarrow 2y = -8 \Rightarrow y = -4$$ (ii) Plot points $(-1,3.5)$, $(2,-1)$, $(4,-4)$ on the graph. (c) Given line AB equation: $3y - 2x = 6$ Solve system: $$\begin{cases} 2y + 3x = 4 \\ 3y - 2x = 6 \end{cases}$$ Multiply first by 3: $$6y + 9x = 12$$ Multiply second by 2: $$6y - 4x = 12$$ Subtract second from first: $$6y + 9x - (6y - 4x) = 12 - 12 \Rightarrow 13x = 0 \Rightarrow x = 0$$ Substitute $x=0$ into $2y + 3(0) = 4$: $$2y = 4 \Rightarrow y = 2$$ Solution: $(0,2)$