Question

For the quadrilateral formed by the lines $4y-3x-1=0,3y+4x+1=0,4y-3x-2=0$ and $3y+4x+2=0$, which among the following statements are true

• A. Quadrilateral formed will be square.
• B. Area of quadrilateral formed is $\frac{1}{25}$ Sq. units.
• C. Slope of diagonals of quadrilateral are $7,\frac{-1}{7}$
• D. Slope of diagonals of quadrilateral are $3,\frac{-1}{3}$

Answer 1

Answer: (A), (B), (C)

Slope of diagonals of quadrilateral are $7,\frac{-1}{7}$

$L_1:-3x+4y-1=0,m_1=\frac{3}{4}$
$L_2:4x+3y+1=0,m_2=-\frac{4}{3}$
$L_3:-3x+4y-2=0,m_3=\frac{3}{4}$
$L_4:4x+3y+2=0,m_1=-\frac{4}{3}$

$\because$ Opposite sides are parallel and adjacent sides are perpendicular. So given quadrilateral can be square or rectangle.
$\because$ distance between $L_1\text{and}{L}_3=$ distance between $L_2\text{and}{L}_4=\frac{1}{5}$ units
$\therefore$ Given quadrilateral represents square, where side length$=\frac{1}{5}$


Area of square$=\frac{1}{25}$ Sq. units
Let slope of diagonal $AC$ be $m$ as it is equally inclined to lines $AB\text{and}AD$
$\Rightarrow \frac{\left(\frac{-4}{3}\right)-m}{1+\left(\frac{-4}{3}\right)\left(m\right)}=-\left(\frac{\left(\frac{3}{4}\right)-m}{1+\left(\frac{3}{4}\right)\left(m\right)}\right)$
$\Rightarrow \frac{4+3m}{3-4m}=\frac{3-4m}{4+3m}$
$\Rightarrow (3m+4{)}^2=(3-4m{)}^2$
$\Rightarrow 3m+4=\pm (3-4m)$
$\Rightarrow 3m+4=3-4m\text{(or)}3m+4=-3+4m$
$\Rightarrow m=-1/7\text{or}m=7$