Through (5,-8); perpendicular to 7y=x-14

Question

Help me 

Answer 1

1. Take the original equation and put it in the form  y=mx+b:

 

          7y=x-14                        original equation

          (7y)/7=(1/7)x-14/7        divide both sides by 7

          y= (1/7)x-2                   simplify

 

2. The slope of the given line is 1/7. The slope of the perpendicular line is the opposite reciprocal of this, or -7.

 

3. Using the point-slope method:

      

         y-y1=m(x-x1)           Point-slope form

         y-(-8)=(-7)(x-5)       (x1,y1)=(5,-8), m=-7

         y+8=(-7)x+35          simplify

         y+8-8=(-7)x+35-8    subtract 8 from both sides

         y=-7x+27                 answer

Answer 2

Hi Kaitlyn,

 

Given the equation of the line as 7y = x-14, my preference is to restate that equation in the slope-intercept form, y=mx+b.  

 

We're almost there, except that the y term has a coefficient of 7, so divide both sides by 7.  This gives us

 

y = x/7 - 2

 

This is important because in this form we can easily see the slope.  For this equation, the slope is 1/7, the coefficient of x.

 

For a line to be perpendicular to another, it's slope must be the negative inverse of the slope of the first line.

 

So the slope of the perpendicular line will be -7, the negative inverse of 1/7.

 

Now we have the slope and a point for the line.

 

We need to use the point-slope form 

 

y-y₁ = m(x-x₁)

 

y-(-8) = (-7)(x-5)

 

Cleaning this up

 

y+8 = -7x+35

 

and finally

 

y = -7x+27