Question

What is the slope of a line that is perpendicular to the lien whose equation is 2y=3x-1

Answer 1

$\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \parallel\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\----------------------\\\\\text{The slope-intercept form:}\ y=mx+b.\\\\\text{We have}\ 2y=3x-1.\ \text{Convert to the slope-intercept form:}\\\\2y=3x-1\qquad\text{divide both sides by 2}\\\\y=\dfrac{3}{2}x-\dfrac{1}{2}\to m_1=\dfrac{3}{2}\\\\\text{Therefore}\\\\m_2=-\dfrac{1}{\frac{3}{2}}=-\dfrac{2}{3}\\\\Answer:\ \boxed{The\ slope\ =\ -\dfrac{2}{3}}$

Answer 2

Answer:

-2/3

Step-by-step explanation:

We are to find the slope of a line which is perpendicular of a line with an equation $2y=3x-1$.

We know that the slope of line which is perpendicular to another line is the negative reciprocal of that perpendicular line.

So writing the given equation of a line in the slope intercept form:

$2y=3x-1$ ---> $y=\frac{3}{2} x-\frac{1}{2}$

Here the slope of this line is $\frac{3}{2}$ so the slope of a line which is perpendicular to the given line will be $-\frac{2}{3}$.