Find an equation for line in y=mx+b format: passes through (2,-1) and is perpendicular to the line 5x+6y=30.?

Question

please include instructions on how to do it if you can.

Answer 1

First find the slope of the other line:

5x+6y=30

y = (-5/6)x + 5

Slope of this line is -5/6.

The slope of the perpendicular will be the negative reciprocal of -5/6 (read: turn -5/6 upside-down and change the sign).

The slope of the perpendicular is 6/5. Now you need to determine its y-intercept using the coordinates of point (2, -1):

b = y - mx

= -1 - (6/5)(2)

= -1 - 12/5

= -17/5

The equation for the perpendicular is

y = (6/5)x - 17/5

http://www.flickr.com/photos/dwread/2827...

Answer 2

seriously? I miss easy math like this.

what grade are you in? this looks like pre-algebra to me. If you need help try a math site but even if you don't get it then, I'll help you out but try it yourself first.

Answer 3

(d) : y=mx+b

(d1) : 5x+6y=30

so (d1) : y = (-5x + 30)/6

(d) perpendicular to (d1) so m.(-5/6) = -1

hence m = 6/5

so we have (d) : y=(6/5)x+b

on the other hand, (d) passes through (2,-1), so if we plug x=2 and y=-1 to the function:

-1 = (6/5).2 + b

we can solve it and have b = -2.2

Conclusion: (d) : y=(6/5)x - 2.2

Really I agree with HP u should consider trying to solve it first before asking