Question

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

Answer 1

Answer:

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

$6x+y=-5$

Step-by-step 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

∴ $y=\dfrac{x}{6}-7$

On Comparing we get

$Slope = m = \dfrac{1}{6}$

The Required line is Perpendicular to the above line.

So,

Product of slopes = - 1

$m\times m_{1}=-1\\Substituting\ m\\ \dfrac{1}{6} m_{1}=-1\\\\m_{1}=-6$

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

$6x+y=-5$