Question

How do you write an equation of a line passing through (0, -9), perpendicular to `7x+6y=5`?

Answer 1

Three steps: 1. find the gradient of the given line, 2. find the gradient of a line perpendicular to it, 3. find the equation of a line with that second gradient that passes through the given point. Solution:
`y=6/7x-9`

Explanation:

Step 1 - find the gradient of the given line:

We need to rearrange it into point-slope form, `y=mx+c` where `m` is the gradient (slope).

`7x+6y=5`

Subtract `7x` from both sides:

`6y = -7x + 5`

Divide both sides by `6`:

`y=-7/6x+5/6`

So the gradient of the given line is `-7/6`.

Step 2 - find the gradient of a line perpendicular to this line:

Call the new gradient `m_1`, then `m_1=-1/m`

That is, to find the gradient of a line perpendicular to another line, we change the sign and take the inverse.

In this case `-1/m=-1/(-7/6)=6/7`

The gradient of the new line we want is `6/7`.

Step 3 - find the equation of the new line:

We know the gradient, and we know that the line passes through the point `(0,-9)`. We can substitute those values into the point-slope form of a line equation:

`y=mx+c`

`-9=6/7(0)+c`

`c=-9`

Therefore the overall equation of the line we want to find is:

`y=6/7x-9`