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Question

# The straight lines 3x+4y=5 and 4xâˆ’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 options are B y+7x=9 D 7y−x=13The lines 3x+4y=5 and 4x−3y=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 4x−3y+5=0is m=4/3±11∓4/3=−7,17 Hence the required equations are y−2=−7(x−1) and y−2=17(x−1) ⇒y+7x=9 and 7y−x=13

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