Question

Area of parallelogram formed by lines
3y - 2x -a, 29-3x ta = 0, 2x-3y +3 a 20
and 3x-2y =o is​

Answer 1

Given:

Lines  3y - 2x -a, 29-3x ta = 0, 2x-3y +3 a 20  and 3x-2y =o

To find:

Area of parallelogram formed by lines  3y - 2x -a, 29-3x ta = 0, 2x-3y +3 a 20 and 3x-2y =o is​

Solution:

Since the given equations are not clear.

I'll be considering an example to show, how to find the area of a parallelogram when given with equations of lines.

3x+4y=7a, 3x+4y=7b, 4x+3y=7c, and 4x+3y=7d

Let,

AB: 3x+4y=7a

CD: 3x+4y=7b

BC: 4x+3y=7c

AD: 4x+3y=7d

The perpendicular distance between AB and CD = d1 = |7b-7a|/√(3²+4²) = 7|b-a|/5

The perpendicular distance between BC and AD = d2 = |7d-7c|/√(3²+4²) = 7|d-c|/5

If θ is the angle between AB and AC,

tan  θ = | m1-m2 / 1+m1m2 |

= | {(-3/4)-(-4/3)} / 1+(-3/4)×(-4/3) |

= 7/24

sin  θ = 7/√(24²+7²) = 7/25

Area of parallelogram = d1d2/sin θ

= [7|b-a|/5 7|d-c|/5] / (7/25)

= 7|b-a||d-c|