Q.The line , for different values of and , passes through the point
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Start your 14-day free trial to unlock the full solution →Group the terms in and : must hold for every , so both brackets vanish; solving the pair gives the fixed point.
The line is really a family of lines, one for each choice of and . A point that lies on every member of the family must satisfy the equation no matter what are, which is only possible if the coefficients of and separately vanish.
Step 1. Regroup by and . Rewrite the equation as
Expand and collect terms in and in :
Step 2. Force independence of . Since and can be chosen independently (they are not proportional to each other), the only way the sum can be zero for every is if each bracket is separately zero:
Step 3. Solve the simultaneous equations. From the first equation, . Substitute into the second: …
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