Question

How do you write an equation of a line passing through (2, 3), perpendicular to `y = (3/2)x − 4`?

Answer 1

We find the line with negative reciprocal slope through the point, which is `y = -2/3 x + 13/3`. Another way to go is to swap the coefficients on `x` and `y` and negate one, and compute the constant from the point, giving `2x + 3y = 13` which is the same line.

Explanation:

When we multiply perpendicular slopes together we get `-1,` meaning perpendicular slopes are the negative reciprocals of each other.

We can read off the slope of `y = 3/2 x - 4` as the coefficient on `x`, which is `3/2`. The slope of the perpendiculars will be the negative reciprocal, `-2/3.`

We fit the constant to the line.

`y = -2/3 x + b`

`3 = -2/3 (2) + b`

`b = 13/3`

So the equation we seek is

`y = -2/3 x + 13/3`

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I prefer not to deal with the slope if I can avoid it. I like first writing the line in standard form:

`- 3x + 2y = -8`

Then the perpendiculars are gotten by swapping the coefficients on `x` and `y`, and negating one. The constant is quickly determined by the point:

`2 x + 3 y = 2(2) + 3(3) = 13`

`2x + 3y = 13`