Question

What is the equation of the line perpendicular to 2x-3y=13 that passes through the point (-6,5)?

Answer 1

The slope-intercept form of a line:

`y=mx+b`

m - slope

b - y-intercept

Convert 2x - 3y = 13 to the slope-intercept form:

`2x-3y=13` subtract 2x from both sides

`-3y=-2x+13` divide both sides by (-3)

`y=(2)/(3)x-(13)/(3)`

Let
`k:y=m_1x+b_1` and
`l:y=m_2x+b_2`.

`l\ \perp\ k\iff m_1m_2=-1\to m_2=-(1)/(m_1)`

We have
`m_1=(2)/(3)` therefore

`m_2=-(1)/((2)/(3))=-(3)/(2)`

Equation of a line:
`y=-(3)/(2)x+b`

Put the coordinates of the point (-6, 5) to the equation of a line:

`5=-(3)/(2)(-6)+b`

`5=(-3)(-3)+b`

`5=9+b` subtract 9 from both sides

`-4=b\to b=-4`

Answer:
`\boxed{y=-(3)/(2)x-4}`