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Determine the slope of a line that is perpendicular to the equation 3x + 6y =18

Sagot :

Medunno13

Rewriting the equation of the given line in slope-intercept form,

[tex]3x+6y=18\\\\x+2y=6\\\\2y=-x+6\\\\y=-\frac{1}{2}x+3[/tex]

This means the slope of the given line is -1/2.

As perpendicular lines have slopes that are negative reciprocals, the answer is 2.

SOLVING

[tex]\Large\maltese\underline{\textsf{A. What is Asked}}[/tex]

Determine the slope of the line perpendicular to 3x+6y=18

[tex]\Large\maltese\underline{\textsf{B. This problem has been solved!}}[/tex]

[tex]\bf{\dfrac{3}{6}x+\dfrac{6}{6}y=\dfrac{18}{6}[/tex] | dividing the ENTIRE equation by 6, to make it easier to write in y=mx+b form

[tex]\bf{\dfrac{1}{2}x+y=3}[/tex] | subtract 1/2 x

[tex]\bf{y=-\dfrac{1}{2}x+3[/tex].

[tex]\cline{1-2}[/tex]

Now, perpendicular lines' slopes are opposite inverses of each other.

The opposite inverse of -1/2 is

                  = 2

[tex]\cline{1-2}[/tex]

[tex]\bf{Result:}[/tex]

                 [tex]\bf{=2}[/tex]

[tex]\LARGE\boxed{\bf{aesthetics\not1\theta l}}[/tex]

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