Grade 12Analytical Geometry

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

Avinash Chothe

7 Years agoGrade 12

1 Answer

Samyak Jain

7 Years ago

2 is a root of ax² + bx + c = 0 i.e. x = 2 is a solution of ax² + bx + c = 0.

$\therefore$ 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 $\Rightarrow$ x + 2(b/a)y + 3(c/a) = 0 … (2)

Substitute value of c/a from (1) in (2). [c/a = – (2b/a + 4)]

$\Rightarrow$ x + 2(b/a)y + 3{– (2b/a + 4)} = 0

x + 2(b/a)y – 6b/a – 12 = 0

(x – 12) + (2b/a)(y – 3) = 0 $\Rightarrow$ (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 = 0

with the parameter t = 2b/a.

$\therefore$ 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).

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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)

Avinash Chothe

1 Answer

Samyak Jain

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