Given the equation of line **e:3x+5y=10**. Determine the slope of the angle, one of the normal vectors and one of the directon vectors. Give the equation of that line passing through the point **P(3;7)** and prependicular to the line **e**.

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First you write the equation in explicit form i.e. `y` on one side and everything else on the other side.

`5y=-3x+10`

`y=-3/5x+2`

The slope is the number next to `x` hence the **slope is**`-3/5.`

If you want to calculate the angle then you have `tan alpha=-3/5=>alpha approx 149^o2' 10''`

Direction vector is the same as for line `y=-3/5x`

because that line is parallel with line `e.` So if we put `x=5` we get `y=-3` hence **direction vector is** `d=[5,-3]`

normal vector is perpendicular to our line `e` , hence parallel to line `y=5/3x` (if you have line `l: y=kx+l` then every line with equation `y=-1/k+m ` is perpendicular to line `l` ) so we put `x=3` and get `y=5` hence **normal vector is** `n=[3,5]` ` `

We already said that perpendicular line has equation `y=5/3x+m` so we only need to find `m` which we do by putting point `P` coordinates instead of `x` and `y` .` `

`7=5/3cdot3+m`

`m=7-5`

`m=2`

**So equation of the line that is perpendicular to `e` and passes through `P` is `y=5/3x+2` .**

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