Question

Find the equation of a line that passes through (0,0) that is perpendicular to 2x+3y=-6

Answer 1

The slope-intercept form of a line:

`y=mx+b`

m - slope

b - y-intercept

Convert 2x + 3y = -6 to the slope-intercet form:

`2x+3y=-6` subtract 2x from both sides

`3y=-2x-6` divide both sides by 3

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

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)`

We have the equation of a line:

`y=(3)/(2)x+b`

Put the coordinates of the point (0, 0) to the equation:

`0=(3)/(2)(0)+b`

`0=b\to b=0`

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