1. **Problem statement:** Find the x-intercept of a line, state the condition for its existence, explain it geometrically, find the gradient by rearranging to gradient-intercept form, state the condition for the gradient to be defined, find gradients of given lines, and explain forms of lines parallel and perpendicular to a given line. 2. **Finding the x-intercept:** The x-intercept is where the line crosses the x-axis, so $y=0$. 3. **Formula:** Substitute $y=0$ into the line equation $Ax + By = C$ to find $x$: $$Ax + B\cancel{0} = C \implies x = \frac{C}{A}$$ 4. **Condition for x-intercept existence:** The x-intercept exists if $A \neq 0$ because division by zero is undefined. 5. **Geometric explanation:** If $A=0$, the line is horizontal ($By=C$), parallel to the x-axis, so it never crosses it, hence no x-intercept. 6. **Gradient-intercept form:** Rearrange $Ax + By = C$ to $y = mx + c$: $$By = C - Ax \implies y = \frac{C}{B} - \frac{A}{B}x$$ So gradient $m = -\frac{A}{B}$. 7. **Condition for gradient to be defined:** $B \neq 0$ because division by zero is undefined. 8. **Find gradients of given lines:** (i) $x + 3y = 2$: $$3y = 2 - x \implies y = \frac{2}{3} - \frac{1}{3}x$$ Gradient $m = -\frac{1}{3}$. (ii) $2x - y = -1$: $$-y = -1 - 2x \implies y = 2x + 1$$ Gradient $m = 2$. (iii) $4x + y = 7$: $$y = 7 - 4x$$ Gradient $m = -4$. (iv) $3x - 2y = -3$: $$-2y = -3 - 3x \implies y = \frac{3}{2} + \frac{3}{2}x$$ Gradient $m = \frac{3}{2}$. (v) $x + 5y = 4$: $$5y = 4 - x \implies y = \frac{4}{5} - \frac{1}{5}x$$ Gradient $m = -\frac{1}{5}$. (vi) $4x - 3y = 12$: $$-3y = 12 - 4x \implies y = -4 + \frac{4}{3}x$$ Gradient $m = \frac{4}{3}$. 9. **Explanation of forms:** (i) Any line parallel to $3x + 5y = 2$ has the same gradient. Gradient is $m = -\frac{3}{5}$. Lines with equation $3x + 5y = d$ have the same $A$ and $B$, so same gradient, differing only in $d$ which shifts the line. (ii) Any line perpendicular to $3x + 5y = 2$ has gradient $m_{\perp} = \frac{5}{3}$ (negative reciprocal of $-\frac{3}{5}$). Lines of form $5x - 3y = d$ have gradient $m = \frac{5}{3}$, so they are perpendicular. **Final answers:** - x-intercept exists if $A \neq 0$. - Gradient defined if $B \neq 0$. - Gradients: (i) $-\frac{1}{3}$, (ii) $2$, (iii) $-4$, (iv) $\frac{3}{2}$, (v) $-\frac{1}{5}$, (vi) $\frac{4}{3}$. - Parallel lines: $3x + 5y = d$. - Perpendicular lines: $5x - 3y = d$.