# Find the line ty=3x-2 when it is perpendicular to the line y=1-x.

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### 2 Answers

You need to convert the equation `ty=3x - 2` to the slope intercept form, such that:

`y = (3/t)x - 2/t => m_1 = 3/t`

You need to use the equation that relates the slopes of orthogonal lines, such that:

`m_1 = -1/m_2`

You need to identify `m_2` as the slope of the orthogonal line, such that:

`y = -x + 1 => m_2 = -1`

Replacing -1 for `m_2` in equation of slopes, yields:

`m_1 = -1/(-1) => {(m_1 = 1),(m_1 = 3/t):} => 3/t = 1 => t = 3`

**Hence, evaluating t, under the given conditions, yields **`t = 3.`

If the 2 given lines are perpendicular, then the product of the values of their slopes is -1.

The given equations are y = 1 - x and ty = 3x - 2, so we'll have to put the equation ty = 3x - 2 in the standard from, which is y = mx+n.

Since y is isolated to the left side, we'll just have to divide by t both sides:

y = (3/t)*x - (2/t)

The other line is:

y = -x+1

So, the slope can be easily determined as m1 = -1.

That means that the slope of the line y = (3/t)*x - (2/t) has the value:

m1*m2 = -1

-1*m2 = -1

-1*(3/t) = -1

t = 3

The line, perpendicular to the line y = 1 - x, is now determined, having as equation: y = x - (2/3).