The slope-intercept form of a line:
[tex]y=mx+b[/tex]
m - slope
b - y-intercept
Convert 2x - 3y = 13 to the slope-intercept form:
[tex]2x-3y=13[/tex] subtract 2x from both sides
[tex]-3y=-2x+13[/tex] divide both sides by (-3)
[tex]y=\dfrac{2}{3}x-\dfrac{13}{3}[/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]
Equation of a line: [tex]y=-\dfrac{3}{2}x+b[/tex]
Put the coordinates of the point (-6, 5) to the equation of a line:
[tex]5=-\dfrac{3}{2}(-6)+b[/tex]
[tex]5=(-3)(-3)+b[/tex]
[tex]5=9+b[/tex] subtract 9 from both sides
[tex]-4=b\to b=-4[/tex]
Answer: [tex]\boxed{y=-\dfrac{3}{2}x-4}[/tex]
