Hi. Please help with this question. Workings would be appreciated. Find the equations of the following line :
perpendicular to 3x-2y =3 and passing through (3,-7)
Note : For perpendicularism , M1 = -1/M2
M2 =-1/M1

or

2x + 3y + 15 = 0

or

2x + 3y = -15

Step-by-step explanation:

We are given an equation in standard form. To find the slope, we can convert it to slope-intercept form, which is y = mx + b.

"m" will be the slope.

Isolate "y":

3x - 2y = 3

-2y = -3x + 3

The slope is 3/2.

m = 3/2

To find the slope of a perpendicular line, find the negative reciprocal by flipping the fraction and changing the negative/positive.

m⊥ = -2/3      which is:

Substitute the coordinates (3, -7) and the slope into slope-intercept form.

y = mx + b                     Start with the general formula.

-7 = -6/3 + b                 Multiply -2/3 by 3

-7 = -2 + b                    Add 2 to both sides to isolate "b"

-5 = b

b = -5                            Keep the variable on the left side

Now we know for our new equation:

m = -2/3

b = -5

Substitute the new information into slope-intercept form.

y = mx + b

                  This is the equation

You might need to convert this is standard form, which is:

ax + by = c or ax + by + c = 0

Use the form where "c" will be positive, or what your teacher prefers.

      Add   to both sides

               Since "c" is negative, add 5 to both sides.

To get rid of the fraction, multiply the whole equation by "3".

3y + 2x + 15 = 0

2x + 3y + 15 = 0           Rearrange to follow ax + by + c = 0

2x + 3y = -15                 Or rearrange to follow ax + by = c

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