Thread: perpendicular passing through a point

1. perpendicular passing through a point

the problem is:

Perpendicuar to y= -5/7x passing through (6/5,0)

I know that to be perp it has to be opp/co so itll be 7/5x but i dont know what to do next at all.

The answer says: y= 7/5x - 6/5

I have no idea how to get there

2. Re: perpendicular passing through a point

Originally Posted by gurrry
the problem is:

Perpendicuar to y= -5/7x passing through (6/5,0)

I know that to be perp it has to be opp/co so itll be 7/5x but i dont know what to do next at all.

The answer says: y= 7/5x - 6/5

I have no idea how to get there
The normal will be of the form y = mx + c, and you have already found m = 7/5

So y = (7/5)x + c

You also know that (x, y) = (6/5, 0) is a point on the line, so you can use this in the equation of your line

0 = (7/5) x (6/5) + c

Solve for c.

3. Re: perpendicular passing through a point

Originally Posted by gurrry
the problem is:
Perpendicuar to y= -5/7x passing through (6/5,0)
I know that to be perp it has to be opp/co so itll be 7/5x but i dont know what to do next at all. The answer says: y= 7/5x - 6/5
You have a point and a slope. Use point-slope form.
$\displaystyle (y-y_0)=m(x-x_0)$
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