The x-intercept of angle bisector of angle between the lines 2x+y−2=0 and 2x+4y+7=0 which contains the fixed point on the family of lines (2cosα+3sinα)x+(3cosα−5sinα)y=5cosα−2sinα for different values of α, is equal to
A
112
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B
−112
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C
−12
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D
12
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Solution
The correct option is A112 Family of lines cosα(2x+3y−5)+sinα(3x−5y+2)=0 will pass through fixed point i.e., point of intersection of 2x+3y−5=0 and 3x−5y+2=0 which is (1,1)
∵2⋅1+1−2>0 and 2⋅1+4⋅1+7>0 ∴ Angle bisector 2x+y−2√5=+(2x+4y+72√5) will contain the point (1,1) ⇒4x+2y−4=2x+4y+7 ⇒2x−2y−11=0