*(-3, -3), perpendicular to y-7/3x+3.*

_{1}=m(x-x

_{1})

I need help on understanding how to find the solution. All answers will be helpful and I appreciate all of you for your explanations.

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Hi Jason;

I am assuming that the equation is...

y=(7/3)x+3

If not, please let me know.

The slope of this line is 7/3.

The slope of the line perpendicular to this is the negative inverse; -3/7.

y=(-3/7)x+b is a formula written in slope-intercept form.

You need point-slope, especially since the y-intercept, value of b, is still unknown.

y-y_{1}=m(x-x_{1})

y--3=(-3/7)(x--3)

Subtracting a negative is the same as adding a positive.

y+3=(-3/7)(x+3)

Let's distribute...

y+3=(-3/7)(x)-(9/7)

Let's subtract 3 from both sides as we proceed to isolate y...

-3+y+3=(-3/7)x-(9/7)-3

y=(-3/7)x-(9/7)-3

Considering the fact that 3=21/7...

y=(-3/7)x-(9/7)-(21/7)

y=(-3/7)x-(30/7)

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

Another method: Be careful writing your questions

Passing through ( -3, -3) , perpendicular to Y = 7/3X + 3 (L1)

Equation of line is always Y = mx + b (1)

Now you have to find m. b from the given information and plug it into (1).

The m is equal to -3/7 ( being perpendicular to line (L1).

so the equation of line is

Y = -3/7X + b

next have to find b, we can plug ( -3,-3) into equation and solve for b

- 3 = -3/7( -3) +b b = -3- 9/7 = -30/7

plugging back into the equation the answer is:

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