Question

Hi. I need to write an equation of a line that contains the given point and is perpendicular to the given line and I don't know how. The ordered pair is (6,4) and the equation is y=3x-2

Answer 1

So the first thing you need is a slope that is perpendicular to the original line. What is the slope of the original line?

Answer 2

I'm not sure. The only numbers that were given to me are the numbers that I gave you.

Answer 3

hint: When you have a linear equation in the form y = mx + b, the slope (m) is the number in front of the x variable.

Answer 4

3

Answer 5

Correct!

Answer 6

Now to find a perpendicular slope, you just take the reciprocal and change the sign:

If m is the slope, then \(\large -\frac{1}{m}\) would be the perpendicular slope. So what is the perpendicular slope to 3?

Answer 7

-(1/3)

Answer 8

Also correct!

Answer 9

Now to find a line with a particular slope m that goes through a particular point (h,k) we have the point slope formula:

\[y-k = m(x-h)\]

In this case we want m to be our perpendicular slope of (-1/3) and our (h,k) point to be (6,4). What will the point slope equation of that look like?

Answer 10

y-4=-(1/3)(x-6)

Answer 11

Exactly.

Answer 12

I'm not sure if that is sufficient for your answer or not, but that is an equation for the line. Sometimes they prefer to have it in y=mx + b form, or some other form.

Answer 13

how would i put it in y=mx+b form?

Answer 14

Solve it for y.

Answer 15

1) distribute the -1/3 to each term in parens
2) add 4 to both sides

Answer 16

would it be y=-(1/3)x+4?

Answer 17

No

Answer 18

You forgot about the (-1/3)(-6)

Answer 19

\(\large y-4 = -\frac{1}{3}(x-6)\)\[\implies y - 4 = -\frac{1}{3}x- (-\frac{1}{3})(6)\]\[\implies y-4 = -\frac{1}{3}x -(-2)\]\[\implies y-4 = -\frac{1}{3}x + 2\]\[\implies y = -\frac{1}{3}x + 6\]