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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 ?

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 ?

Grade:12

3 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
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+\lambda(bx - 2ay - 3a)=0...............(1)
(a+b\lambda)x+(2b-2a\lambda)y+3b-3\lambdaa=0
As the line is parallel to x axis
a+b\lambda=0 so \lambda=(-b/a)
Putting \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
askIITians faculty
Latika Leekha
askIITians Faculty 165 Points
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
askIITians Faculty
Gowtham
19 Points
7 years ago
The straight line ax+by+c=0, where abc not equal to zero ,will pass through the first quardrant if 1) ac>0,bc>0 2)c>0 and bc0 and/ or ac>0. 4) ac

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