The lines (a+b)x+(a−b)y−2ab=0,(a−b)x+(a+b)y−2ab=0 and x+y=0 forms an isosceles triangle whose vertical angle is
A
π2
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B
|tan−1(2aba2−b2)|
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C
|tan−1(ab)|
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D
2|tan−1(ab)|
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Solution
The correct option is B|tan−1(2aba2−b2)| For (a+b)x+(a−b)y−2ab=0 Slope is m1=−a+ba−b And for (a−b)x+(a+b)y−2ab=0 Slope is m2=−a−ba+b Then angle is tanθ=∣∣∣m1−m21+m1m2∣∣∣=∣∣
∣
∣
∣∣−a+ba−b+a−ba+b1+a+ba−ba−ba+b∣∣
∣
∣
∣∣=∣∣∣a2+b2−2ab−a2−b2−2aba2−b2+a2−b2∣∣∣=∣∣∣2aba2−b2∣∣∣