# Find the equation of the line perpendicular to the line -3x+y=30 and passing through the point (-9,2)

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The equation of the line perpendicular to the line -3x + y = 30 and passing through the point (-9, 2) has to be determined.

-3x + y = 30

=> y = 3x + 30

This line has a slope of 3, the slope of a line perpendicular to this line is -1/3.

As the line passes through (-9, 2), the equation of the line is (y - 2)/(x + 9) = -1/3

=> 3(y - 2) = -(x + 9)

=> 3y - 6 = -x - 9

=> x + 3y + 3 = 0

**The equation of the required line is x + 3y + 3 = 0**

The line is -3x+y=30

>y=3x+30

therefore slope(m) is 3 by the formula y=mx+c

since we have to find the equation of the perpendicular line:

let its slope be (m1)

we know in case of perpendicula lines> m*m1=-1

therefore m1=-1/3

now,by the formula y-y1=m(x-x1)

the equation is> y-2=-1/3(x-(-9)) [as it passes through the point (-9,2)

>3(y-2)=-1(x+9)

>3y-6=-x-9

>x+3y=-3

**Therefore, the req. equation is x+3y=-3**