# find the equation of the line that is perpendicular to the line y=1/9x + 8 and contains the point (-6,0) y=

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Find the equation of a line that is perpendicular to `y=1/9x+8` containing the point (-6,0):

The slope of the given line is m=1/9 The slope of a line perpendicular to this line is -9 (slopes of perpendicular lines are opposite reciprocals, or their product is (-1))

Thus we have a slope of -9 and a point (-6,0).

The equation of the line can be found using the point-slope form If given m and a point `(x_1,y_1)` the equation is `y-y_1=m(x-x_1)`

`y-0=-9(x-(-6))`

`y=-9(x+6)`

`y=-9x-54`

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**The line perpendicular to `y=1/9x+8` ` `going through (-6,0) is `y=-9x-54` **

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The graphs:

Gradient of the given line = m1=1/9

If the gradient of the line we need = m2

Since these two lines are perpendicular;

m1*m2=-1

1*m2/9=-1

m2=-9

Equation is given by;

y-y0=m(x-x0)

y-0=-9(x-(-6))

y+9x+54=0