Exercises: Finding line equations

$P(2|0);Q(-2|2)$

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

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

Plug $m$ and the coordinates of a point, e.g. $P(2|0)$ into the general line equation and solve for $t$.

$0=-\frac{1}{2}\cdot 2+t|+\frac{1}{2}\cdot 2$

$t=1$

Substitute $m$ and $t$ into the general line equation.

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

Drawing the line

$P(0.5|1.5);Q(5|3)$

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

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

Plug $m$ and the coordinates of a point, e.g. $Q(5|3)$ into the general line equation and solve for $t$.

$3=\frac{1}{3}\cdot 5+t|-\frac{1}{3}\cdot 5$

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

Substitute $m$ and $t$ into the general line equation.

$\Rightarrow$ $y=\frac{1}{3}x+\frac{4}{3}$

$P(-2|1)$; $Q(6|4)$

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

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

Plug $m$ and the coordinates of a point, e.g. $P(-2|1)$ into the general line equation and solve for $t$.

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

$1=-\frac{6}{8}+t$

$1=-\frac{3}{4}+t|+\frac{3}{4}$

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

Substitute $m$ and $t$ into the general line equation.

$\Rightarrow$ $y=\frac{3}{8}x+\frac{7}{4}$

$P(-4|1)$; $Q(1|-1)$

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

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

Plug $m$ and the coordinates of a point, e.g. $P(-4|1)$ into the general line equation and solve for $t$.

$1=\frac{8}{5}+t|-\frac{8}{5}$

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

Substitute $m$ and $t$ into the general line equation.

$\Rightarrow$ $y=-\frac{2}{5}x-\frac{3}{5}$