Question

Find the distance of the point (2,5) from the line through (10,4) which points in the direction of −6I−2J.

Answer 1

Can you find the equation of the line through (10,4) with slope 1/3?

Answer 2

is that what it is asking? The wording is confusing

Answer 3

Yes, it is a little odd. All we get from "in the direction of" is slope information.

Answer 4

so the slope is
-2/-6?

Answer 5

Or, 1/3. Let's see what you get.

Answer 6

what should
I do with the (2,5) info?

Answer 7

You'll need the line, first. Put it in NORMAL form.

Answer 8

Ok.. I have no idea how to do this..

Answer 9

Come on! You can do it.

y = mx + b, right?

Slope is 1/3

y = (1/3)x + b

Point is (10,4)

4 = (1/3)10 + b ==> b = 4 - 10/3 = 2/3

y = (1/3)x + 2/3 <== Point-Slope Form

3y = x + 2

x - 3y + 2 = 0 <== Normal Form

Now, all we have to do is calculate the distance from this line to the point (2,5). Any ideas?

Answer 10

From where on this line though?

Answer 11

straight perpendicular?

Answer 12

That is an excellent question. Exactly the question that needed to be asked.

You can go about it two ways. One that requires the greatest attention.

1) Find the equation of the line perpendicular to x - 3y + 2 = 0 through (2,5).
2) Calculate the point of intersection.
3) Use the distance formula to calculate the distance.

The other way is why I wanted it in Normal Form.

Given x - 3y + 2 = 0 and (2,5), we have \(\dfrac{|(2) - 3(5) + 2|}{\sqrt{1^{2}+3^{2}}{}}\) -- Done!

Answer 13

Oh duh..
Thanks!