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Show that the straight lines given by `x(a+2b)+y(a+3b)=a` for different values of `a` and `b` pass through a fixed point.

A

`(2,1)`

B

`(2,-1)`

C

`(-2,1)`

D

None of these

Text Solution

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The correct Answer is:
B
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If a + 2b + 3c = 0 " then " a/3+(2b)/3+c=0 and comparing with line ax + by + c, we get x = 1/3 & y = 2/ 3 so there will be a point (1/3,2/3) from where each of the lines of the form ax + by + c = 0 will pass for the given relation between a,b,c . We can say if there exists a linear relation between a,b,c then the family of straight lines of the form of ax + by +c pass through a fixed point . If a , b,c are in A.P., then the line ax + 2by + c = 0 passes through

If a + 2b + 3c = 0 " then " a/3+(2b)/3+c=0 and comparing with line ax + by + c, we get x = 1/3 & y = 2/ 3 so there will be a point (1/3,2/3) from where each of the lines of the form ax + by + c = 0 will pass for the given relation between a,b,c . We can say if there exists a linear relation between a,b,c then the family of straight lines of the form of ax + by +c pass through a fixed point . If a , b, c are consecutive odd integers then the line ax + by + c = 0 will pass through

If the algebraic sum of perpendiculars from n given points on a variable straight line is zero then prove that the variable straight line passes through a fixed point

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