Lines are drawn parallel to the line 4x – 3y + 2 = 0, at a distance {3 \over 5} from the origin. Then which one of the following points lies on any of these lines ?

The region represented by| x – y | $$\le$$ 2 and | x + y| $$\le$$ 2 is bounded by a :

If the two lines x + (a – 1) y = 1 and 2x + a2y = 1 (a$$\in$$R – {0, 1}) are perpendicular, then the distance of their point of intersection from the origin is :

Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is :

Suppose that the points (h,k), (1,2) and (–3,4) lie on the line L1 . If a line L2 passing through the points (h,k) and (4,3) is perpendicular to L1 , then k \over h equals :

If the system of linear equations x – 2y + kz = 1 2x + y + z = 2 3x – y – kz = 3 has a solution (x,y,z), z $$\ne$$ 0, then (x,y) lies on the straight line whose equation is :

Let O(0, 0) and A(0, 1) be two fixed points. Then the locus of a point P such that the perimeter of $$\Delta$$AOP is 4, is :

A point on the straight line, 3x + 5y = 15 which is equidistant from the coordinate axes will lie only in :

If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, $$\beta$$), then $$\beta$$ equals :

If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is :

Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant :

Two sides of a parallelogram are along the lines, x + y = 3 & x – y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is :

If the line 3x + 4y – 24 = 0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is :

A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R (3, – 2) are fixed points, then the locus of the centroid of $$\Delta$$PQR is a line :

If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r cannot be equal to :

Let the equations of two sides of a triangle be 3x $$-$$ 2y + 6 = 0 and 4x + 5y $$-$$ 20 = 0. If the orthocentre of this triangle is at (1, 1), then the equation of its third side is :

If a straight line passing through the point P(–3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is :

Two vectors $$\overrightarrow{A}$$ and $$\overrightarrow{B}$$ have equal magnitudes. The magnitude of $$\left(\overrightarrow{A}+\overrightarrow{B}\right)$$ is 'n' times the magnitude of $$\left(\overrightarrow{A}-\overrightarrow{B}\right)$$ . The angle between $$\overrightarrow{A}$$ and $$\overrightarrow{B}$$ is -

Let $$\left|\overrightarrow{A_1}\right|=3$$, $$\left|\overrightarrow{A_2}\right|=5$$ and $$\left|\overrightarrow{A_1}+\overrightarrow{A_2}\right|=5$$. The value of $$\left(2\overrightarrow{A_1}+3\overrightarrow{A_2}\right)\left(3\overrightarrow{A_1}-\overrightarrow{2A_2}\right)$$ is :-

In the cube of side ‘a’ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be -