[3x-2y+3=0],[4x+3y-47=0]

Prove that the area of the parallelogram formed by the lines 3x-4y+a=0,\ 3x-4y+3a=0,\ 4x-3y-a=0\ a n d \4x-3y-2a =0\ i s (2a^2)/7s qdotu n i t sdot

Prove that the area of the parallelogram formed by the lines 3x-4y+a=0,\ 3x-4y+3a=0,\ 4x-3y-a=0 and 4x-3y-2a=0 is (2a^2)/7 sq. units

The area of the parallelogram formed by the lines 4y-3x-a=0, 3y-4x+a=0, 4y-3x-3a=0, 3y-4x+2a=0 is

Given the four lines with the equations x+2y-3=0 , 3x+4y-7=0 , 2x+3y-4=0 , 4x+5y-6=0 , then

Given the four lines with the equations x+2y-3=0 , 3x+4y-7=0 , 2x+3y-4=0 , 4x+5y-6=0 , then

Given the four lines with equations x+2y-3=0, 3x+4y-7=0, 2x+3y-4=0 and 4x+5y-6=0 , then:

Find the orthocenter of the triangle formed by the lines x + 2y = 0, 4x + 3y - 5 =0, 3x + y = 0 .

Prove that the area of the parallelogram contained by the lines 4y-3x-a=0,3y-4x+a=0,4y-3x-3a=0, and 3y-4x+2a=0 is (2/7)a^2dot

Prove that the area of the parallelogram contained by the lines 4y-3x-a=0,3y-4x+a=0,4y-3x-3a=0, and 3y-4x+2a=0 is (2/7)a^2dot