Q. 95.0( 2 Votes )

# If the diagonals of the quadrilateral formed by the lines l_{1}x + m_{1}y + n_{1} = 0, l_{2}x + m_{2}y + n_{2} = 0, l_{1}x + m_{1}y + n’_{1} = 0 and l_{2}x + m_{2}y + n_{2}’ = 0 are perpendicular, then write the value of l^{12} – l_{2}^{2} + m^{1}_{2} – m^{2}_{2}.

Answer :

__Given:__

l_{1}x + m_{1}y + n_{1} = 0 … (1)

l_{2}x + m_{2}y + n_{2} = 0 … (2)

l_{1}x + m_{1}y + n’_{1} = 0 … (3)

l_{2}x + m_{2}y + n_{2}‘ = 0 … (4)

__To find:__

The value of l^{12} – l_{2}^{2} + m^{1}_{2} – m^{2}_{2}.

**Explanation:**

Assuming:

(1), (2), (3) and (4) represent the sides AB, BC, CD and DA, respectively.

The equation of diagonal AC passing through the intersection of (2) and (3) is given by

l_{1}x + m_{1}y + n’_{1}+ λ(l_{2}x + m_{2}y + n_{2}) = 0

⇒ (l_{1} + λl_{2})x + (m_{1} + λm_{2})y + (n_{1}’ + λn_{2}) = 0

⇒ Slope of diagonal AC

Also, the equation of diagonal BD, passing through the intersection of (1) and (2), is given by

l_{1}x + m_{1}y + n_{1} + μ(l_{2}x + m_{2}y + n_{2}) = 0

⇒ l_{1} + μl_{2}x + m_{1} + μm_{2}y + n_{1} + μn_{2} = 0

⇒ Slope of diagonal BD

The diagonals are perpendicular to each other.

∴

⇒ (l_{1} + λl_{2})(l_{1} + λl_{2}) = (-m_{1} + λm_{2})(m_{1} + μm_{2})

Let λ = -1, μ = 1

⇒ (l_{1} – l_{2})(l_{1} + l_{2}) = (-m_{1}-m_{2})(m_{1} + m_{2})

⇒ (l_{1}^{2}-l_{2}^{2)} = (-m_{1}^{2}-m_{2}^{2)}

⇒ (l_{1}^{2}-l_{2}^{2}) + (m_{1}^{2}-m_{2}^{2}) = 0

Hence, (l_{1}^{2}-l_{2}^{2}) + (m_{1}^{2}-m_{2}^{2}) = 0

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