Question

Write and graph an equation that is perpendicular to the equation x-3y=6 and through point (2,-3)

Answer 1

Answer:

y = mx + b   is slope-intercept form of the linear equation.

y - y1 = m(x - x1)   is point-slope form of the linear equation.

Ax + By = C  (A ≥ 0)   is standard form of the linear equation.

~~~~~~~~~~~

....

              1  

y – 5  =  ——  (x + 1)

              3  

         1             1

y  =  —— x  +  ——  +  5

         3             3  

                   1             16

(1)   y  =  ——  x  +  ——

                  3              3  

Step-by-step explanation:

Answer 2

Answer:

The answer to your question is     y = -3x + 3

Step-by-step explanation:

See the graph below

Data

Perpendicular to x - 3y = 6

Point = (2, -3)

Process

1.- Find the slope of the line given

         x - 3y = 6

             -3y = -x + 6

                y = -1/-3x + 6/-3

                y = 1/3x - 2

slope = 1/3

2.- Find the slope of the new line

          slope = -3

3.- Find the equation of the new line

         y + 3 = -3(x - 2)

         y + 3 = -3x + 6

         y = -3x + 6 - 3

         y = -3x + 3