Are the following lines parallel, perpendicular, or niether?

Question

L1 with equation x - 7y = 14
L2 with equation 7x + y = 7

Answer 1

Lines that are parallel have the same slope (sign and value)
m1 = m2
Lines that are perendicular have negative reciprocal slopes
m1 = -1/m2

x - 7y = 14 rearrange to this form y = mx + b
-7y = -x + 14
y = +1/7x - 2 (slope, m1 is a +1/7)

7x + y = 7 rearrange to this form y = mx + b
y = -7x + 7 (slope, m2 is a -7)

So they are perpendicular

Answer 2

For two lines to be parallel then their gradient must be equal.
For two lines to be perpendicular then the product of their gradients must be -1.
L1: x - 7y = 14
= 7y = -14 + x
Thus y = -2 + x/7
The gradient of L1 is the co-efficient of x which is +1/7.
L2: 7x + y = 7
= y = 7 -7x
The gradient of L2 is the coefficient of x which is -7
The product of the two co-efficients is -1
So, the two lines are perpendicular.

Answer 3

They are perpendicular:

In slope-intercept form, L1 is y=x/7-2 and has a slope of 1/7.
Conversely, L2 is y=-7x+7, and has a slope of -7.

-7 and 1/7 are negative reciprocals of each other.

Answer 4

As has already been posted, the slopes of these
2 lines are negative reciprocals, so they are
perpendicular.
Here is another way to do this:
Look at the direction vectors for each line and
take their dot product. If it is 0, that means the
lines are perpendicular.
The direction vector for the first line is (1,-7)
The direction vector for the second line is (7,1).
Their dot product is 7-7 = 0.
Thus the lines are perpendicular.

Answer 5

Write them in the form y = mx+n

L1 y=1/7x-2

L2 y= -7x+7 The slopes (m) are negative inverses so the lines are perpendicular

Answer 6

Neither, the slopes are not the same, nor are they opposite reciprical.

Answer 7

neither