# write an equation in slope intercept form for: a line through (6,1) and perpendicular to the line y=-1/3x-4

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You should remember the relation between the slopes of two orthogonal lines such that: `m_1*m_2=-1` .

Notice that the slope intercept form of the line `y=(-1/3)x - ` 4 gives the slope `m_1 = -1/3` , hence `m_2 = -1/(-1/3) =gt m_2 = 3` .

You need to write the slope intercept form of the line that passes through (6,1), with `m_2=3` such that:

`y - 1 = 3(x - 6)`

Opening the brackets yields:

`y - 1 = 3x - 18 =gt y = 3x - 18 + 1`

`y = 3x - 17`

**Hence, the slope intercept form of equation of the line, orthogonal to `y=(-1/3)x - 4` is `y = 3x - 17` .**

a line through (6,1) and perpendicular to the line y=-1/3x-4

The slope of the original line is -1/3. The slope of a line perpendicular to this is the negative reciprocal: 3

We want a line that goes through the points (6,1) with slope of 3. Writing this in point-slope form is easiest.

y - y1 = m(x - x1)

y - 1 = 3(x - 6)

To get this into slope-intercept form, just solve for y and make sure to distribute the slope.

y = 3x - 18 + 1

y = 3x - 17