Question

How do you write an equation of a line passing through (0, 2), perpendicular to `y = 4x – 3`?

Answer 1

`x+4y-8=0`

Explanation:

The slope of line: `y=4x-3\equiv y=mx+c` is `4`.

Hence the slope `m` of the line perpendicular to the given line: `y=4x-3`

`m=-1/{4}`

Now, the equation of line passing through the point `(x_1, y_1)\equiv(0, 2)` & having a slope `m=-1/4` is given as

`y-y_1-m(x-x_1)`

`y-2=-1/4(x-0)`

`4y-8=-x`

`x+4y-8=0`

Answer 2

`y=-(1/4)x+2`

Explanation:

First, I'd start with the slope of the new equation. We know that this unknown equation is perpendicular to `y=4x-3`. Since it is perpendicular, the slope of the equation has to be the reciprocal and negative, giving us the new slope of `-1/4`.

So to recap, so far we have `y=-(1/4)x+b`, with `b` representing the `y`-intercept.

We know the `y`-intercept is `2` because `(0,2)` is a coordinate on the `y`-axis, but I will also solve for it, so it is more clear.

We can find the `y`-intercept by plugging in the coordinates that we are given `(0,2)` into this new equation and solving for `b`.

`2=-(1/4)(0)+b`

`2=b`

Now that we have the `y`-intercept, we can put together an equation.
`y=-(1/4)x+2`