# perpendicular passing through a point

• September 18th 2011, 07:55 AM
gurrry
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
• September 18th 2011, 08:04 AM
Prove It
Re: perpendicular passing through a point
Quote:

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.
• September 18th 2011, 08:06 AM
Plato
Re: perpendicular passing through a point
Quote:

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.
$(y-y_0)=m(x-x_0)$