Question

The point of intersection of diagonals of the parallelogram formed by the lines 3x−2y+5=0, x+3y-5=0, 3x−2y+11=0 and x+3y+3=0 is

• A. (−2, −1)
• B. (−2, 1)
• C. (2, −1)
• D. (2, 1)

Answer 1

The correct option is B

(−2, 1)

Line Midway between the lines 3x - 2y + 5 = 0 . 3x - 2y + 11 = 0 is

3x - 2y + $\frac{5+11}{2}=0$ $\Rightarrow$ 3x - 2y + 8 = 0 .............. (1)

Similarly line Midway between the lines x + 3y - 5 = 0 , x + 3y + 3 = 0 is

x + 3y + $\left(\frac{-5+3}{2}\right)=0\Rightarrow$ x + 3y - 1 = 0 ...... (2)

point of intersection of (1) & (2) is

Same as points of intersection of diagonals

$\therefore$ Required point is (-2,1)

Note :- Don't try finding diagonals and solving them ( very lenghty - time tabing)