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`        If 2 is a root of ax2+bx+c=0 then point of con- currence of lines ax+2by+3c=0 is1) (12,3)     2) (4,2)     3) (1,2)     4) (2, 3)`
3 months ago

## Answers : (1)

```							2 is a root of ax2 + bx + c = 0 i.e. x = 2 is a solution of ax2 + bx + c = 0.  a(2)2 + b(2) + c = 0  i.e.   4a + 2b + c = 0    or   4 + 2b/a + c/a = 0     …. (1)Now, ax + 2by + 3c = 0     x + 2(b/a)y + 3(c/a)  =  0           … (2)Substitute value of c/a from (1) in (2).    [c/a = – (2b/a + 4)]x + 2(b/a)y + 3{– (2b/a + 4)} = 0x + 2(b/a)y – 6b/a – 12 = 0(x – 12) + (2b/a)(y – 3) = 0      (x – 12) + t (y – 3) = 0   …. (3),   where t = 2b/a.Equation (3) represents a family of lines of  x – 12 = 0  and  y – 3 = 0with the parameter t = 2b/a. Their point of concurrence is given by x – 12 = 0 and y – 3 = 0  i.e.  x = 12 and y = 3.Thus, the required point of concurrence is (12,3).
```
3 months ago
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