Exercises: Linear functions and line equations

Determine the equation of the line $g$ that is parallel to the line $h$ and passes through the point $P$ .

h: $y = 3 x - 2$ ; $P (1 | 0) $

For this task you need the following basic knowledge: Slope/Gradient of a line
$y = 3 x - 2$ ; $P (1 | 0)$
For the line to be parallel to $h$ , it must have the same gradient/slope.
The gradient of a line is called $m$ in the general line equation.
$m = 3$
Setting up the equation
Plug $m = 3$ and the point $P (1 | 0) = (x | y)$ into the general line equation.
$0$ $=$ $3 \cdot 1 + t$ $- 3$
↓
solve for $t$
$t$ $=$ $- 3$
Substitute $m$ and $t$ into the general line equation.
$\Rightarrow$ $y = 3 x - 3$
$h$ : $y = x - 4$ ; $P (1 | 2) $

For this task you need the following basic knowledge: Slope/Gradient of a line
$y = x - 4$ ; $P (1 | 2)$
For the line to be parallel to $h$ , it must have the same gradient/slope.
The gradient of a line is called $m$ in the general line equation.
$m = 1$
Setting up the equation
Plug $m = 1$ and the point $P (1 | 2) = (x | y)$ into the general line equation.
$2$ $=$ $1 + t$ $- 1$
↓
solve for $t$
$t$ $=$ $1$
Substitute $m$ and $t$ into the general line equation.
$\Rightarrow$ $y = x + 1$
$h$ : $y = 4 x$ ; $P (5 | 18) $

For this task you need the following basic knowledge: Slope/Gradient of a line
$y = 4 x$ ; $P (5 | 18)$
For the line to be parallel to $h$ , it must have the same gradient/slope.
The gradient of a line is called $m$ in the general line equation.
$m = 4$
Setting up the equation
Plug $m = 4$ and the point $P (5 | 18) = (x | y)$ into the general line equation.
$18$ $=$ $4 \cdot 5 + t$ $- 20$
↓
solve for $t$
$t$ $=$ $- 2$
Substitute $m$ and $t$ into the general line equation.
$\Rightarrow$ $y = 4 x - 2$
$h$ : $y = - 2 x + 1$ ; $P (- 1 | 4)$

For this task you need the following basic knowledge: Slope/Gradient of a line
$y = - 2 x + 1$ ; $P (- 1 | 4)$
For the line to be parallel to $h$ , it must have the same gradient/slope.
The gradient of a line is called $m$ in the general line equation.
$m = - 2$
Setting up the equation
Plug $m = - 2$ and the point $P (- 1 | 4) = (x | y)$ into the general line equation.
$4$ $=$ $- 2 \cdot (- 1) + t$ $- 2$
↓
solve for $t$
$t$ $=$ $2$
Substitute $m$ and $t$ into the general line equation.
$\Rightarrow$ $y = - 2 x + 2$

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$0$	$=$	$3 \cdot 1 + t$	$- 3$
	↓	solve for $t$
$t$	$=$	$- 3$

$2$	$=$	$1 + t$	$- 1$
	↓	solve for $t$
$t$	$=$	$1$

$18$	$=$	$4 \cdot 5 + t$	$- 20$
	↓	solve for $t$
$t$	$=$	$- 2$

$4$	$=$	$- 2 \cdot (- 1) + t$	$- 2$
	↓	solve for $t$
$t$	$=$	$2$

Exercises: Linear functions and line equations

Setting up the equation

Setting up the equation

Setting up the equation

Setting up the equation