Question

If the area of the parallelogram whose sides are x + 2y + 3 = 0, 3x + 4y - 5 = 0, 2x+4y+5=0 and 3x + 4y - 10 = 0 is 'a' sq. unit. Find the value of 4a

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Answer 1

Given lines are x+2y+3=0
$y=\frac{-x}{2}-\frac{3}{2}.......\left(1\right)$
3x+4y-5=0
$y=\frac{-3}{4}x+\frac{5}{4}=0......\left(2\right)$
2x+4y+5=0
$y=\frac{-x}{2}-\frac{5}{4}......\left(3\right)$
3x+4y-10=0
$y=\frac{-3}{4}x+\frac{5}{2}......\left(4\right)$
slope of line 1 and line 3 is same. So these lines are parallel.
slope of line 2 and line 4 is same. So these lines are parallel.

Area of the parallelogram bounded by the lines
$y=m_{1}x+c_1,y=m_{1}x+c_2,y=m_{2}x+d_1,m_{2}x+d_2$ is given by $|\frac{(c_1-c_2)(d_1-d_2)}{m_1-m_2}|$

So, area of parallelogram = $\left|\frac{(\frac{-3}{2}+\frac{5}{4})(\frac{5}{2}-\frac{5}{4})}{(\frac{-1}{2}+\frac{3}{4})}\right|$
$=\left|\frac{(-\frac{1}{4})\left(\frac{5}{4}\right)}{\left(\frac{1}{4}\right)}\right|$={5\over 4}\) sq.unit
$a=\frac{5}{4}$ sq. unit
$4a=4\times \frac{5}{4}=5$ sq. unit