$P(2\mathrm{\mid }0);Q(-2\mathrm{\mid }2)P(2|0);\; Q(-2|2)$

Determine the gradient $mm$ of the general line equation $y=m\cdot x+ty=m\backslash cdot\; x+t$ using the difference quotient.

$m=\frac{2-0}{-2-2}=\frac{2}{-4}=-\frac{1}{2}m=\backslash frac\{2-0\}\{-2-2\}=\backslash frac2\{-4\}=-\backslash frac12$

Plug $mm$ and the coordinates of a point, e.g. $P(2\mathrm{\mid }0)P(2|0)$ into the general line equation and solve for $tt$.

$0=-\frac{1}{2}\cdot 2+t\mid +\frac{1}{2}\cdot 20=-\backslash frac\{1\}\{2\}\backslash cdot2+t\backslash \; \backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash left|\{+\backslash frac12\backslash cdot2\}\backslash right.$

$t=1t=1$

Substitute $mm$ and $tt$ into the general line equation.

$\Rightarrow \backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;$ $y=-\frac{1}{2}x+1y=-\backslash frac12x+1$

Drawing the line

$P(0.5\mathrm{\mid }1.5);Q(5\mathrm{\mid }3)P(0.5|1.5);\; Q(5|3)$

Determine the gradient $mm$ of the general line equation $y=m\cdot x+ty=m\backslash cdot\; x+t$ using the difference quotient.

$m=\frac{3-1.5}{5-0.5}=\frac{1.5}{4.5}=\frac{1}{3}m=\backslash frac\{3-1.5\}\{5-0.5\}=\backslash frac\{1.5\}\{4.5\}=\backslash frac13$

Plug $mm$ and the coordinates of a point, e.g. $Q(5\mathrm{\mid }3)Q(5|3)$ into the general line equation and solve for $tt$.

$3=\frac{1}{3}\cdot 5+t\mathrm{\mid }-\frac{1}{3}\cdot 53=\backslash frac13\backslash cdot5+t\; \backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;|\{-\backslash frac13\backslash cdot5\}$

$t=\frac{4}{3}t=\backslash frac43$

Substitute $mm$ and $tt$ into the general line equation.

$\Rightarrow \backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;$ $y=\frac{1}{3}x+\frac{4}{3}y=\backslash frac13x+\backslash frac43$

$P(-2\mathrm{\mid }1)P(-2|1)$; $Q(6\mathrm{\mid }4)Q(6|4)$

Determine the gradient $mm$ of the general line equation $y=m\cdot x+ty=m\backslash cdot\; x+t$ using the difference quotient.

$m=\frac{4-1}{6-(-2)}=\frac{3}{8}m=\backslash frac\{4-1\}\{6-(-2)\}=\backslash frac38$

Plug $mm$ and the coordinates of a point, e.g. $P(-2\mathrm{\mid }1)P(-2|1)$ into the general line equation and solve for $tt$.

$1=\frac{3}{8}\cdot (-2)+t1=\backslash frac38\backslash cdot(-2)+t$

$1=-\frac{6}{8}+t1=\; -\backslash frac68\; +t$

$1=-\frac{3}{4}+t\mathrm{\mid }+\frac{3}{4}1=-\backslash frac34+t\; \backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;|+\backslash frac3\; 4$

$t=\frac{7}{4}t=\backslash frac74$

Substitute $mm$ and $tt$ into the general line equation.

$\Rightarrow \backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;$ $y=\frac{3}{8}x+\frac{7}{4}y=\backslash frac38x+\backslash frac74$

$P(-4\mathrm{\mid }1)P(-4|1)$; $Q(1\mathrm{\mid }-1)Q(1|-1)$

Determine the gradient $mm$ of the general line equation $y=m\cdot x+ty=m\backslash cdot\; x+t$ using the difference quotient.

$m=\frac{-1-1}{1-(-4)}=-\frac{2}{5}m=\backslash frac\{-1-1\}\{1-(-4)\}=-\backslash frac25$

Plug $mm$ and the coordinates of a point, e.g. $P(-4\mathrm{\mid }1)P(-4|1)$ into the general line equation and solve for $tt$.

$1=\frac{8}{5}+t\mathrm{\mid }-\frac{8}{5}1=\backslash frac85+t\; \backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;|-\; \backslash frac\; 8\; 5$

$t=-\frac{3}{5}t=-\backslash frac35$

Substitute $mm$ and $tt$ into the general line equation.

$\Rightarrow \backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;$ $y=-\frac{2}{5}x-\frac{3}{5}y=-\backslash frac25x-\backslash frac35$