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Perpendicular line?

I've tried for about an hour but never get it, i just know that they should be equal to -1.

A(3,4),B(5,8). A line perpendicular to AB passes through the origin and through the point (6,r). Find r.

I've tried y=mx+b but nothing, can anyone help?
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Avatar for Zacken
Zacken
22
Original post by amorae
I've tried for about an hour but never get it, i just know that they should be equal to -1.

A(3,4),B(5,8). A line perpendicular to AB passes through the origin and through the point (6,r). Find r.

I've tried y=mx+b but nothing, can anyone help?


What is the gradient of AB?

The gradient of the new line is (-1/gradient of AB).

So since you know two points on the perpendicular line (6, r) and (0,0)

Then (r-0) / (6 - 0) = -1/gradient of AB.
Original post by amorae
I've tried for about an hour but never get it, i just know that they should be equal to -1.

A(3,4),B(5,8). A line perpendicular to AB passes through the origin and through the point (6,r). Find r.

I've tried y=mx+b but nothing, can anyone help?
You use [(y2-y1)/(x2-x1)] to get the gradient of AB - let's call this m. Which is one and which is two doesn't really matter. Then do you know how to find the perpendicular gradient?
Original post by Zacken
Then (r-0) / (6 - 0) = -1/gradient of AB.
I've never thought of it in that way (all at once) before.
What topic does this come under?
Seen this come up a few times would be good if I could learn it.
Reply 5
Avatar for Zacken
Zacken
22
Original post by junayd1998
What topic does this come under?
Seen this come up a few times would be good if I could learn it.


Coordinate geometry, equations of lines?

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