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=-(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=(3)/(2)x-(1)/(2)\to m_1=(3)/(2)\\\\\text{Therefore}\\\\m_2=-(1)/((3)/(2))=-(2)/(3)\\\\Answer:\ \boxed{The\ slope\ =\ -(2)/(3)}`

Answer 2

-2/3

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=(3)/(2) x-(1)/(2)`

Here the slope of this line is
`(3)/(2)` so the slope of a line which is perpendicular to the given line will be
`-(2)/(3)`.