Question

What is an equation of the line that is perpendicular to y-4=2(×-6) and passes thru the point (-3,-5)

Answer 1

An equation of the line that is perpendicular to y-4=2(x-6) and passes thru the point (-3,-5) will be:

• 
  `y=-(1)/(2)x-(13)/(2)`

Explanation:

We know the point-slope form of the line equation is

`y-y_1=m\left(x-x_1\right)`

where

• m is the slope of the line

• (x₁, y₁) is the point

Given the line

y-4 = 2(x-6)

comparing with the point-slope form of the line equation

The slope = m = 2

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:

slope = m = 2

so, the slope of the line perpendicular to y-4 = 2(x-6) will be:

– 1/m = -1/2 = -1/2

substituting the values of the slope = -1/2 and the point (-3, -5)

`y-y_1=m\left(x-x_1\right)`

`y-\left(-5\right)=-(1)/(2)\left(x-\left(-3\right)\right)`

`y+5=-(1)/(2)\left(x-\left(-3\right)\right)`

`y+5=-(1)/(2)\left(x+3\right)`

Subtract 5 from both sides

`y+5=-(1)/(2)\left(x+3\right)`

Simplify

`y=-(1)/(2)x-(13)/(2)`

Therefore, an equation of the line that is perpendicular to y-4=2(x-6) and passes thru the point (-3,-5) will be:

• 
  `y=-(1)/(2)x-(13)/(2)`