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What is an equation of the line that passes through the point (-2,7) and is perpendicular to the line x-6y=42 ?

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Answer:

The equation of the line that passes through the point (-2,7) and is perpendicular to the line x-6y=42 is


6x+y=-5

Explanation:

Given:

Let,

point A( x₁ , y₁) ≡ ( -2 , 7)

To Find:

Equation of Line that passes through the point (-2,7) and is perpendicular to the line x-6y=42=?

Solution:


x-6y=42 ..................Given

which can be written as


y=mx+c

Where m is the slope of the line


y=(x)/(6)-7

On Comparing we get


Slope = m = (1)/(6)

The Required line is Perpendicular to the above line.

So,

Product of slopes = - 1


m* m_(1)=-1\\Substituting\ m\\ (1)/(6) m_(1)=-1\\\\m_(1)=-6

Slope of the required line is -6

Equation of a line passing through a points A( x₁ , y₁) and having slope m is given by the formula,

i.e equation in point - slope form


(y-y_(1))=m(x-x_(1))

Now on substituting the slope and point A( x₁ , y₁) ≡ ( -2, 7) and slope = -6 we get


(y-7)=-6(x--2)=-6(x+2)=-6x-12\\\\\therefore 6x+y=-5.......is\ the required\ equation\ of\ the\ line

The equation of the line that passes through the point (-2,7) and is perpendicular to the line x-6y=42 is


6x+y=-5

User Skytiger
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