Question

If the diagonals of the quadrilateral formed by the lines $l-1x+m_{1}y+n_1=0,l_{2}x+m_{2}y+n_2=0,l_{1}x+m_{1}y+n_1^{\prime }=0$ and $(l_{2}x+m_{2}y+n_2^{\prime }=0$are perpendicular,then write the value of $l_1^2-l_2^2+m_2-m_2^2$

Answer 1

The given lines are

$l-1x+m_{1}y+n_1=0$ ...(i)

$l_{2}x+m_{2}y+n_2=0$ ...(ii)

$l_{1}x+m_{1}y+n_1^{\prime }=0$ ...(iii)

$(l_{2}x+m_{2}y+n_2^{\prime }=0$ ...(iv)

Let (i), (ii),(iii) and (iv) represent the sides AB,BC,CD and DA,rewpectively.

The equation of diagonal AC passing through the intersection of (ii) and (iii) is given by

$l_{1}x+m_{1}y+n^{\prime }+\lambda (l_{2}x+m_{2}y+n_2)=0$

$(l_1+\lambda l_2)x+(m_1+\lambda m_2)y+(n_1^{\prime }+\lambda n_2)=0$

$\Rightarrow$ Slope of diagonal AC =- $\left(\frac{l_1+\lambda n_2}{m_1+\lambda m_2}\right)$

Also, the equation of diagonal BD,passing through the intersention of (i) and (ii) ,is given by

$l_{1}x+m_{1}y+n_1+\mu (l_{2}x+m_{2}y+n_2)=0$

$\Rightarrow$ Slope of diagonal BD=-$\left(\frac{l_1+\mu l_2}{m_1+\mu m_2}\right)$

The diagonal are perpendicular to each other

$\therefore \left(\frac{l_1+\lambda n_2}{m_1+\lambda m_2}\right)\left(\frac{l_1+\mu l_2}{m_1+\mu m_2}\right)$=-1

$\Rightarrow (l_1+\lambda l_2)(l_1+\mu l_2)=-(m_1+\lambda m_2)(m_1+\mu m_2)$

$\Rightarrow (l_1-l_2)(l_1+l_2)=-(m_1-m_2)(m_1+m_2)$

$\Rightarrow (l_1^2-l_2^2)=-(m_1^2-m_2^2)$

$\Rightarrow (l_1^2-l_2^2)+(m_1^2-m_2^2)=0$