# Determine m

• March 30th 2009, 12:56 AM
beq!x
Determine m
THere are given lines p1: $\left( {m - 6} \right)x + \left( {m - 1} \right)y + 13 = 0$ and p2: $\left( {2m + 1} \right)x + 3my - 9 = 0$
Determine $m$ so that lines p1 and p2 have one common point F and F is a point of this line q: $2x+y-5$
• March 30th 2009, 04:39 AM
HallsofIvy
Quote:

Originally Posted by beq!x
THere are given lines p1: $\left( {m - 6} \right)x + \left( {m - 1} \right)y + 13 = 0$ and p2: $\left( {2m + 1} \right)x + 3my - 9 = 0$
Determine $m$ so that lines p1 and p2 have one common point F and F is a point of this line q: $2x+y-5$

2x+ y- 5 isn't a line. Do you mean 2x+ y- 5= 0?

For the first line, y= -[(m-6)/(m-1)]x- 13/(m-1) (as long as m-1 is not 0). For the second line, y= -[(2m+1)/2m]x+ 3/m. At any point where they intersect y= -[(m-6)/(m-1)]x- 13/(m-1)= -[(2m+1)/2m]x+ 3/m. Solve that for x, in terms of m, of course, and use either of the two equaitions to determine y. Finally, put those values of x and y into 2x+ y- 5= 0 to get a single equation for m.
• March 30th 2009, 12:54 PM
beq!x
yes i mean 2x+ y- 5= 0
but why we have to do this: y= -[(m-6)/(m-1)]x- 13/(m-1)= -[(2m+1)/2m]x+ 3/m ?
and i tried in this way but i didn't come to these solutions m=2 and m=2.8 that are in the book.
im confused with this :S
• March 30th 2009, 10:44 PM
earboth
1. Solve p1 for y
2. Solve p2 for y
3. If the lines intersect then the y-values must be equal.
4. You'll get HallsofIvy's equation with a slight change:
-[(m-6)/(m-1)]x- 13/(m-1)= -[(2m+1)/(3m)]x+ 3/m

5. Proceed as HallofIvy has suggested.
• April 1st 2009, 09:22 AM
beq!x
thank you guys :D