Q. 95.0( 2 Votes )

# If the diag

Given:

l1x + m1y + n1 = 0 … (1)

l2x + m2y + n2 = 0 … (2)

l1x + m1y + n’1 = 0 … (3)

l2x + m2y + n2‘ = 0 … (4)

To find:

The value of l12 – l22 + m12 – m22.

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
l1x + m1y + n’1+ λ(l2x + m2y + n2) = 0

(l1 + λl2)x + (m1 + λm2)y + (n1’ + λn2) = 0

Slope of diagonal AC Also, the equation of diagonal BD, passing through the intersection of (1) and (2), is given by
l1x + m1y + n1 +
μ(l2x + m2y + n2) = 0

l1 + μl2x + m1 + μm2y + n1 + μn2 = 0

Slope of diagonal BD The diagonals are perpendicular to each other. (l1 + λl2)(l1 + λl2) = (-m1 + λm2)(m1 + μm2)

Let λ = -1, μ = 1

(l1 – l2)(l1 + l2) = (-m1-m2)(m1 + m2)

(l12-l22) = (-m12-m22)

(l12-l22) + (m12-m22) = 0

Hence, (l12-l22) + (m12-m22) = 0

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