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Question
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that `1/p^2 = 1/a^2 + 1/b^2`.
Solution
It is known that the equation of a line whose intercepts on the axes are a and b is
`x/a + y/b = 1`
or bx + ay = ab
or bx + ay - ab = 0 ......(1)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = `|Ax_1 + By_1 + C|/sqrt(A^2 + B^2)`.
On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = b, B = a, and C = -ab
Therefore, if p is the length of the perpendicular from point (x1, y1) = (0, 0) to line (1), we obtain
`p = |A(0) + B(0)-ab|/sqrt(b^2 + a^2)`
= `p = |-ab|/sqrt(a^2 + b^2)`
On sqauring both sides, we obtain
`p^2 = (-ab)^2/(a^2 + b^2)`
= p2 (a2 + b2) = a2b2
= `(a^2 + b^2)/(a^2b^2) = 1/(p^2)`
= `1/p^2 = 1/a^2 + 1/b^2`.
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