Question

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

Answer 1

`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(x-x_1)`

Plugging the parallel slope, `m=1` and the point `(-4,1)` gives us

`y-1=1(x-(-4))`

`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 `(-4,1)`

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

Answer 2

`y=x+5`

Explanation:

A line parallel to a given line say `ax+by+c=0` is always `ax+by+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 `(-4,1)`, 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 `ax+by+c=0` is of the type `bx-ay+k=0`. Observe that while coefficients of `x` and `y` are interchanged, sign of only one of them is changed.