# How do you write the equation of the line that is parallel to y=x+3 and passes through (-4,1)?

Jun 16, 2017

$y = x + 5$

#### Explanation:

If one line is parallel to another line, then they have the same slope, $m$, as found in the point slope form

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

Plugging the parallel slope, $m = 1$ and the point $\left(- 4 , 1\right)$ gives us

$y - 1 = 1 \left(x - \left(- 4\right)\right)$

$y - 1 = x + 4$

Adding to $1$ both sides

$y = x + 5$

Graphing them both at the same time, reveals two separate graphs which have the same slope (i.e., they are parallel). The top graph passes through $\left(- 4 , 1\right)$

graph{(y-x-3)(y-x-5)=0}

Jun 16, 2017

$y = x + 5$

#### Explanation:

A line parallel to a given line say $a x + b y + c = 0$ is always $a x + b y + k = 0$, where $k$ is another constant. Note that coefficients and signs of $x$ and $y$ remain same,, while constant is different.

Hence a line parallel to $y = x + 3$ will have the equation

$y = x + k$ and as it passes through $\left(- 4 , 1\right)$, we have

$1 = - 4 + k$ or $k = 4 + 1 = 5$

Hence equation of desired line is $y = x + 5$

Note $-$ Equation of line perpendicular to $a x + b y + c = 0$ is of the type $b x - a y + k = 0$. Observe that while coefficients of $x$ and $y$ are interchanged, sign of only one of them is changed.