Home
Class 12
MATHS
The point of concurrence of the line ax+...

The point of concurrence of the line ax+by+c=0 and a,b,c satisfy the relation 3a+2b+4c=0 is

A

(3/2,1/4)

B

(3/4,1/4)

C

(3/4,1/2)

D

(3/2,1/2)

Text Solution

Verified by Experts

The correct Answer is:
C
Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

If sin theta and cos theta are the roots of the equation ax^(2)-bx+c=0, then a,b and c satisfy the relation

The coefficients of the quadratic equation ax^(2)-bx+c=0 satisfy the relation 4a +c -2b lt 0 and c gt 0 then it has a root in the interval

Knowledge Check

  • If sin theta and costheta are the roots of the equation ax^2+bx+c=0 , then a,b and c satisfy the relation

    A
    `a^b+b^2+2ac=0`
    B
    `a^b-b^2+2ac=0`
    C
    `a^b+c^2+2ab=0`
    D
    `a^b-b^2-2ac=0`

  • If a point satisfying the line ax+by=c, then

    A
    It will lie on the line
    B
    It will lle in upper half plane
    C
    It will lle in lower half plane
    D
    None of the above

  • 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

    A
    (0,0)
    B
    (1,-2)
    C
    (-1,1)
    D
    (1,-1)

    • Similar Questions

    Explore conceptually related problems

    For a>b>c>0, if the distance between (1,1) and the point of intersection of the line ax+by-c=0 and bx+ay+c=0 is less than 2sqrt(2) then,(A)a+b-c>0(B)a-b+c 0(D)a+b-c<0

    Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4a x+""2a y+c=""0 and 5b x+""2b y+d=""0 lies in the fourth quadrant and is equidistant from the two axes then (1) 2b c-3a d=""0 (2) 2b c+""3a d=""0 (3) 3b c-2a d=""0 (4) 3b c+""2a d=""0

    The coefficient of the equation ax^(2)=bx+c=0 where a!=0 , satisfy the inequality (a+b+c)(4a-2b+c)lt0 . Prove that this equation has 2 distinck real solutions.

    If A(-1, 0), B(1, 0) and C(3, 0) are three given points, then the locus of point D satisfying the relation DA^2 + DB^2 = 2DC^2 is (A) a straight line parallel to x-axis (B) a striaght line parallel to y-axis (C) a circle (D) none of these

    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

    Home
    Profile