Question

How do you write an equation of a line passing through (5, 8), perpendicular to `y = 2/3 x - 3`?

Answer 1

`y=-3/2x+31/2`

Explanation:

First find the slope of the perpendicular line:

The slope of a perpendicular line is defined as `m_|_=-1/m`

We know that the slope of the given line is `2/3` so the slope of the perpendicular line is then:

`-1/(2/3)` or `-1*3/2=-3/2`

Next, we can use the point slope formula: `y-y_1=m(x-x_1)` to find the equation of the line by substituting the perpendicular slope and the point `(5,8)` for `(x_1,y_1)` Thus:

`y-8=-3/2(x-5)`

We can express this in `y=mx+b` form is desired by simply solving for `y`:

`y-8=-3/2x+15/2`

`ycancel(-8+8)=-3/2x+15/2+8`

*Find the LCD for `15/2` and `8` which is `2`

`y=-3/2x+15/2+8/1(2/2)`

`y=-3/2x+15/2+16/2`

`y=-3/2x+31/2`

The graph of the two lines will then look like this: graph{(y-2/3x+3)(y+3/2x-31/2)=0 [-1.92, 18.08, -0.32, 9.68]}