-3x-6y=17;(6,3)

Question

write the equation of a line that is perpendicular to the given line that passes through  the given point.

Answer 1

The equation of the line is given in Standard Form:   Ax + By = C:      -3x-6y=17

The slope of the line is (-A/B), which, here is (-1/2).

The slope of all perpendicular lines is the negative reciprocal of the slope of this line.

      That is, (-1/slope), which is (B/A) or (2).

 

The equation of the line in Standard Form so far is   2x - y = D     (where D ≠ C)

  So, we need point (6,3) to determine D.

        2(6) - (3) = D

             9 = D

 

The perpendicular line has the equation:  2x - y = 9

  In slope-intercept form:  y = 2x + 9

 

Note:  to find the answer in slope-intercept form, you may want to first convert the equation of the line:

        -3x - 6y = 17     (original line)

               -6y = 3x + 17        (add 3x to both sides)

               y = -x/2 - 17/6      (divide both sides by -6)

      

The perpendicular line has slope of 2; use the point (6,3)

        y = 2x + b      (b is the y-intercept)

        3 = 2(6) + b

        -9 = b

The equation of the perpendicular line (in Slope-Intercept form) is:    y = 2x -9

 

(Note:  THX, I corrected some lines)