Question

 

Given four lines with equations x + 2y – 3 = 0, 3x + 4y – 7 = 0,

2x + 3y – 4 = 0, 4x + 5y – 6 = 0, then

Solution

Correct option is

None of these

 

The point of intersection of

       x + 2y – 3 = 0 and 2x + 3y – 4 = 0

is found to be (–1, 2), The coordinates of this point satisfy the equation 4x + 5y – 6 = 0 but not 3x + 4y – 7 = 0. Hence all the four lines are not concurrent. Again since three of the lines meet at a point they cannot from a quadrilateral.

SIMILAR QUESTIONS

Q1

Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are

Q2

If a, b, c are all unequal and different from one and the points  are collinear then ab + bc + ca =

Q3

Consider three points

              ,

               and

              , then

Q4

The lines p(p2 + 1)x – y + q = 0 and (p2 + 1)2 x + (p2 + 1)y + 2q = 0 are perpendicular to a common line for

Q5

The number of integral values of m, for which the x-coordinates of the point of intersection of the lines 2x + 4y = 9 and 

y = mx + 1 is also an integral is

Q6

Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) andR(7, 3). The equation of the line passing through

(1, –1) and parallel toPS is

Q7

Let PQR be a right angled isosceles triangle, right angled at P(2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is

Q8

The point (2, 1) is shifted through a distance  units measured parallel to the line x + y = 1 in the direction of decreasing ordinates to reach Q.The image of Q w.r.t. given line is

Q9

Given the family of lines a(2x + y + 4) + b(x – 2y – 3) = 0. The number of lines belonging to the family at a distance  from any point (2, –3) is

Q10

Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nxand y = nx + 1 equals.