find the value of k if the straight line 2x+3Y+4+k(6x-y+12)=0 is perpendicular to the line 7x+5y-4=0
2007-04-21 18:42:32 · 4 answers · asked by cutegirl 3 in Science & Mathematics ➔ Mathematics
The two given lines are perpendicular if their slopes are negative reciprocals of each other.
2x + 3y + 4 + k(6x - y + 12) = 0
7x + 5y - 4 = 0
Take the second line.
7x + 5y - 4 = 0
5y = -7x + 4
y = (-7/5)x + 4/5
The slope m = -7/5.
The slope of the first line must be the negative reciprocal of this m' = 5/7.
Take the first line.
2x + 3y + 4 + k(6x - y + 12) = 0
2x + 3y + 4 + 6kx - ky + 12k = 0
(2 + 6k)x + (3 - k)y + (4 + 12k) = 0
(3 - k)y = - (2 + 6k)x - (4 + 12k)
y = - [(2 + 6k)/(3 - k)]x - (4 + 12k)/(3 - k)
m' = - (2 + 6k)/(3 - k) = 5/7
-7(2 + 6k) = 5(3 - k)
-14 - 42k = 15 - 5k
-37k = 29
k = -29/37
2007-04-23 22:30:17 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
Just to confirm a previous answer. I got -29/37 as well.
2007-04-22 02:02:04 · answer #2 · answered by Josh G 2 · 0⤊ 0⤋
I think it's -29/37
2007-04-22 01:50:09 · answer #3 · answered by princess 2 · 0⤊ 0⤋
errr...never came across dis qns in ma maths lesson. wen u say perpendicular..is the equation equal? hahah..dunno dun care
2007-04-22 01:48:50 · answer #4 · answered by Anonymous · 0⤊ 3⤋