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find the slope of the line which is perpendicular to the line 7 x + 11 Y - 2=0​

Sagot :

gmany

Answer:

[tex]\huge\boxed{slope=-\dfrac{7}{11}}[/tex]

Step-by-step explanation:

METHOD 1

The slope-intercept form of an equation of a line:

[tex]y=mx+b[/tex]

[tex]m[/tex] - slope

[tex]b[/tex] - y-intercept

We have the equation in the general form [tex](Ax+By+C=0)[/tex].

Convert to the slope-intercept form:

[tex]7x+11y-2=0[/tex]    subtract 7x from both sides

[tex]11y-2=-7x[/tex]     add 2 to both sides

[tex]11y=-7x+2[/tex]     divide both sides by 11

[tex]y=-\dfrac{7}{11}x+\dfrac{2}{11}[/tex]

METHOD 2

The general form of an equation of a line:

[tex]Ax+By+C=0[/tex]

The slope:

[tex]m=\dfrac{-A}{B}[/tex]

We have

[tex]7x+11y-2=0\\\\A=7;\ B=11;\ C=-2[/tex]

Substitute:

[tex]m=\dfrac{-7}{11}=-\dfrac{7}{11}[/tex]

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