Question

What is the an equation of the line that goes through (−1, −3) and is perpendicular to the line `2x + 7y + 5 = 0`?

Answer 1

`y = 7/2 x + 1/2`

Explanation:

Let's write the equation of the line `2x + 7y + 5 = 0` in standard form first.
To do so, solve the equation for `y`:

`color(white)(xxx) 2x + 7y + 5 = 0`

`<=> 7y = -2x - 5`

`<=> y = -2/7 x - 5/7`

Now, we know that `m_1 = -2/7` is the slope of this line.

To make the line that you need to construct perpendicular to the given line, its slope `m` must fulfill the condition `m = -1/m_1`, so the slope of the line that you need to create is

`m = 7/2`.

Now that you have the slope `m` and your point `(-1, -3)`, you can compute the equation of the line.

The standard form is

`y = mx + n`.

Since you know that the point `(-1, -3)` is on your line, `x = -1` and `y = -3` must fulfill this equation.
Thus, plug `x`, `y` and `m` and solve for `n`:

`-3 = 7/2 * (-1) + n`

`<=> n = 1/2`

The equation of your line is `y = 7/2 x + 1/2`.