So, the slope of a line perpendicular to the line whose equation is 4x+6y=108 is: 3/2
Further explanation:
Given equation of line is:
[tex]4x+6y=108[/tex]
In order to find the slope of given line, we have to convert it to point slope from for that we have to isolate y on one side.
[tex]4x+6y=108\\2(2x+3y)=108\\2x+3y=\frac{108}{2}\\2x+3y=54\\3y= -2x+54\\\frac{3y}{3}=\frac{-2x+54}{3}\\y=-\frac{2}{3}x+\frac{54}{3}\\y=-\frac{2}{3}x+18[/tex]
Let m1 be the slope of given line
and m2 be the slope of the line perpendicular to given line
Then
[tex]m_1=-\frac{2}{3}[/tex]
The product of slopes of two perpendicular lines is -1
[tex]m_1.m_2=-1\\-\frac{2}{3}.m_2= -1\\m_2 = -1 * -\frac{3}{2}\\m_2=\frac{3}{2}[/tex]
So, the slope of a line perpendicular to the line whose equation is 4x+6y=108 is: 3/2
Keywords: Slope, Slope of perpendicular lines
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