Find the equation of the line that contains the point (-1, 3) and is perpendicular to the line 2x + 4y = -1?

Question

I am not really sure how to do this.

Answer 1

Find the equation of the line that contains the point (-1, 3) and is perpendicular to the line 2x + 4y = -1.

Calculate the slope m, of the given line.

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

m = -1/2

The slope slope m' of the line perpendicular to the given line is the negative reciprocal.

m' = -1/m = 2

Now write the equation of the line thru (-1,3) that is perpendicular to the given line.

y - 3 = 2(x + 1)
y - 3 = 2x + 2
y = 2x + 5

Answer 2

Slope form of the given line 2x + 4y = -1 is:

y = - ½ x - 1/4. So that its slope is -½ .

Any perpendicular to the line 2x + 4y = -1 will have slope "m" such that

m*(-½ ) = - 1,

giving m = 2.

Now any line containing the point (-1, 3) and having slope m = 2 will be:

y - 3 = 2* (x +1) i.e. 2x - y + 5 = 0.

Answer 3

Rewrite any linear equation to the form y = mx + c
Line perpendicular to this one will have slope -1/m

2x + 4y = -1
is
y = (-x/2) - (1/4)
Then slope is -1/2

But the line is perpendicular to this one then the slope is -1/(-1/2) = 2

As line passes through (-1,3)

y = 2x + c, substitute (-1,3)
3 = -2 + c so c = 5

Line is y = 2x + 5