The Equation To The Side Of A Triangle Are x – 3y = 0, 4x + 3y = 5 And 3x + Y = 0. The Line 3x – 4y = 0 Passes Through 

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

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

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

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

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

The area bounded by the curves 

A line making an angle  with the + ive direction of x-axis passes throughP(5, 6) to meet the line x = 6 at Q and y = 9 at R the QR is

A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. Then the point Odivides the segment PQ in the ratio

Orthocenter of triangle whose vertices are (0, 0), (3, 4), (4, 0) is

The three lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15 from the sides of a triangle. The line (1, 2) is its

One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is