1. **Problem statement:** (i) Show that two distinct lines $y = ax + c$ and $y = a'x + c'$ have a common point unless $a = a'$. Also show this when one line has infinite slope. (ii) Deduce that the unique line through a point $P$ parallel to a line $\ell$ has the same slope as $\ell$. (iii) Find the equation of the line through $P = (2,2)$ parallel to $y = 3x$. 2. **Step (i): Common point of two lines** Two lines $y = ax + c$ and $y = a'x + c'$ intersect if there exists $(x,y)$ satisfying both equations. Set them equal: $$ax + c = a'x + c'$$ Rearranged: $$ax - a'x = c' - c$$ $$ (a - a')x = c' - c $$ If $a \neq a'$, solve for $x$: $$ x = \frac{c' - c}{a - a'} $$ Substitute back to find $y$: $$ y = a \cdot \frac{c' - c}{a - a'} + c $$ Thus, lines intersect at one point. If $a = a'$, then: $$ (a - a')x = 0 = c' - c $$ If $c' \neq c$, no solution (parallel lines). If $c' = c$, lines coincide. For infinite slope (vertical line), equation is $x = k$. A line $y = ax + c$ intersects $x = k$ at: $$ y = a k + c $$ So they intersect at $(k, a k + c)$. 3. **Step (ii): Parallel line through point $P$** Two lines are parallel if they have the same slope. Given line $\ell: y = ax + c$ and point $P = (x_0, y_0)$, the unique line through $P$ parallel to $\ell$ has slope $a$. Equation: $$ y - y_0 = a(x - x_0) $$ 4. **Step (iii): Equation of line through $P=(2,2)$ parallel to $y=3x$** Slope $a = 3$. Using point-slope form: $$ y - 2 = 3(x - 2) $$ Simplify: $$ y - 2 = 3x - 6 $$ $$ y = 3x - 4 $$ **Final answer:** The line through $(2,2)$ parallel to $y=3x$ is: $$ y = 3x - 4 $$