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

2 Answers

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

Jul 21, 2018

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