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Question

Let a,b,c,d be non-zero numbers. If point of intersection of 6ax+2ay+c=0 and 5bx+3by+d=0 lies in second quadrant such that its distance from x-axis is twice its distance from y-axis then

A
7bc+10ad=0
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B
bc+2ad+=0
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C
7ad+10bc=0
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D
bc2ad=0
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Solution

The correct option is A bc+2ad+=0
Let the coordinates of the intersection point in the second quadrant be (α,2α) lies on both lines

6ax+2ay+c=0 and 5bx+3by+d=0

The coordinates of the intersection point are represented as (α,2α) because it is given that the distance from x-axis twice the distance from the y-axis

These coordinates should satisfy the equations of the intersecting lines

6a(α)+2a(2α)+c=0

6aα+4aα+c=0

c=2aα

α=c2α....(1)

From the other equation

5b(α)+3b(2α)+d=0

5bα+6bα+d=0

bα=d

α=da.....(2)

Equating (1) and (2)

c2a=db

bc=2ad

2ad+cb=0

bc+2ad=0

option b is the correct answer


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