The slope-intercept form of a line:
[tex]y=mx+b[/tex]
m - slope
b - y-intercept
Convert 2x + 3y = -6 to the slope-intercet form:
[tex]2x+3y=-6[/tex] subtract 2x from both sides
[tex]3y=-2x-6[/tex] divide both sides by 3
[tex]y=-\dfrac{2}{3}x-2[/tex]
Let [tex]k:y=m_1x+b_1[/tex] and [tex]l:y=m_2x+b_2[/tex]
[tex]l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}[/tex]
We have [tex]m_1=-\dfrac{2}{3}[/tex]
Therefore
[tex]m_2=-\dfrac{1}{-\frac{2}{3}}=\dfrac{3}{2}[/tex]
We have the equation of a line:
[tex]y=\dfrac{3}{2}x+b[/tex]
Put the coordinates of the point (0, 0) to the equation:
[tex]0=\dfrac{3}{2}(0)+b[/tex]
[tex]0=b\to b=0[/tex]
Answer: [tex]\boxed{y=\dfrac{3}{2}x}[/tex]
