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

$y=-\dfrac{2}{3}x-2$

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

$l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}$

We have $m_1=-\dfrac{2}{3}$

Therefore

$m_2=-\dfrac{1}{-\frac{2}{3}}=\dfrac{3}{2}$

We have the equation of a line:

$y=\dfrac{3}{2}x+b$

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

$0=\dfrac{3}{2}(0)+b$

$0=b\to b=0$

Answer: $\boxed{y=\dfrac{3}{2}x}$