# The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0, where (a, b) ? (0, 0) is ?

SHAIK AASIF AHAMED
9 years ago
Hello student,
The line passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0 is
ax + 2by + 3b+$\dpi{80} \lambda$(bx - 2ay - 3a)=0...............(1)
(a+b$\dpi{80} \lambda$)x+(2b-2a$\dpi{80} \lambda$)y+3b-3$\dpi{80} \lambda$a=0
As the line is parallel to x axis
a+b$\dpi{80} \lambda$=0 so $\dpi{80} \lambda$=(-b/a)
Putting $\dpi{80} \lambda$=(-b/a) in equation (1) we get
ax + 2by + 3b+(-a/b)(bx - 2ay - 3a)=0
y(2b+2a2/b)+3b+3a2/b=0
on simplifying we get y=-3/2
so it is 3/2 units below x-axis
Thanks and Regards
Shaik Aasif
Latika Leekha
9 years ago
The given lines are ax + 2by + 3b = 0 and bx – 2ay – 3a = 0.
Then the required equation is of the form
ax + 2by + 3b + l(bx – 2ay – 3a) = 0.
(a + bl)x + (2b – 2al)y + (3b-3a) = 0
Since, the lline is given to be parallel to x-axis, hence we must have the coefficient of x to be zero.
Hence, a+bl = 0
l = -a/b
So, the given equation reduces to (2b + 2a2/b)y + (3b + 3a2/b) = 0
This yields 2y + 3 = 0, which is a line parallel to x-axis and since, y = -3/2, so it lies below the x-axis at a distnce of 3/2 from the origin.
Thansk & Regards
Latika Leekha