Question

find the equation of a perpendicular line that passes through the point (3,3/2) Given the line 6x-4y=12

Answer 1

Let's rewrite the equation of the line in the form y = mx + b.
6x - 12 = 4y

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

Since
`m_t = (3)/(2)`,

`m_n = (-2)/(3)`

Now, we have a point and a slope, so we'd just substitute everything to the point-gradient form.


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

`3y - (9)/(2) = -2(x - 3)`

`3y - (9)/(2) = -2x + 6`

`2x + 3y - (9)/(2) - 6 = 0`

`2x + 3y - (21)/(2) = 0`

`4x + 6y - 21 = 0`


`6y = 21 - 4x`

`y = -(2x)/(3) + (7)/(2)`