write an equation of the line that is perpendicular to y=-2/7x+9 and that passes through (4,-6)

Question

i would like to know the answer to this problem please and thank you. if anyone knows please let me know id appreciate it.

Answer 1

I am going to assume you mean y = (2/7)x + 9, i.e. a straight line, not an hyperbola.

 

The slope of the perpendicular line is the negative reciprocal of the slope of the line; in this case the slope is -7/2

 

Hence, -7/2 = (y+6)/(x-4).

 

I will leave it to you to get it into standard form.

Answer 2

The slope of the perpendicular line is 7/2

 

(4,-6) ---> X=4 and y=-6

 

B = y - mx = (-6) - (7/2)(4)

                = -6 - 14

               = -20

 

y = (7/2)X - 20

Answer 3

The given line:      y=-(2/7)x+9     has a slope of (-2/7).

  All perpendicular lines have a slope of (7/2).  That is the negative reciprocal of (-2/7).

 

The point-slope form of the equation of the perpendicular line that goes through (4,-6) is:

      y-(-6) = (7/2)(x-4)

 

     2(y+6) = 7(x-4)

      2y + 12 = 7x - 28

The Standard Form of the equation of the perpendicular line is:

      7x - 2y = 40

 

     2y - 7x - 40

The slope-intercept form of the equation of the perpendicular line is:

     y = (7/2)x - 20

 

Check:

    Does     (7/2)4  - 20 = (-6)   ?

                   14 - 20 = - 6   ?

                       - 6 = - 6    ?    yes.  The perpendicular line intersects at (4,-6)