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Question

The straight lines 3x+4y=5 and 4x−3y=15 intersect at the point A. On these lines points B and C are chosen so that AB=AC.Possible equation(s) of the line BC passing through (1,2) is/are

A
x+7y=9
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B
y+7x=9
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C
7xy=13
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D
7yx=13
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Solution

The correct option is D 7yx=13
The lines 3x+4y=5 and 4x3y=15 respectively.
Hence the two given straight lines are at right angles.

So, according to the condition AB=AC there are 2 such lines are possible which make an angle of 45° with both lines
Now slope of the line which makes an angle of 45° with 4x3y+5=0is
m=4/3±114/3=7,17
Hence the required equations are
y2=7(x1) and y2=17(x1)
y+7x=9 and 7yx=13

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