Question

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​

Answer 1

`y = -(2)/(3)x - 5`

or

2x + 3y + 15 = 0

or

2x + 3y = -15

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

`y = (-3)/(-2)x + (3)/(-2)`

`y = (3)/(2)x - (3)/(2)`

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:
`-(2)/(3)`

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

y = mx + b Start with the general formula.

`-7 = -(2)/(3)(3) + b`

-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

`y = -(2)/(3)x - 5` 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.

`y = -(2)/(3)x - 5`

`y + (2)/(3)x= -(2)/(3)x - 5 + (2)/(3)x` Add
`(2)/(3)x` to both sides

`y + (2)/(3)x= - 5`

`y + (2)/(3)x +5= - 5 + 5` Since "c" is negative, add 5 to both sides.

`y + (2)/(3)x + 5= 0`

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

`3y + (3)((2)/(3)x) + (3)(5)= 0`

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