Latika Leekha
Last Activity: 10 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
askIITians Faculty
