If P is the length of perpendicular from the origin to the line whose intercept on the Axes are A and B then show that one upon 1/p^2=1/a^2 1/b^2?

Question

Answer

Answer

Answer

Concept of Line Intercepts
The equation of a line with intercepts $ A $ and $ B $ on the x-axis and y-axis can be expressed as:
$$
\frac{x}{A} + \frac{y}{B} = 1
$$
This implies that the line intersects the x-axis at $ (A, 0) $ and the y-axis at $ (0, B) $.
Finding the Length of the Perpendicular (P)
The length of the perpendicular $ P $ drawn from the origin $ (0, 0) $ to the line can be calculated using the formula:
$$
P = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}
$$
In our line equation, rearranging gives us $ Ax + By - C = 0 $ where $ C = AB $. Thus, we have:
$$
P = \frac{|0 + 0 - AB|}{\sqrt{B^2 + A^2}} = \frac{AB}{\sqrt{A^2 + B^2}}
$$
Establishing the Relationship
We want to show that:
$$
\frac{1}{P^2} = \frac{1}{A^2} + \frac{1}{B^2}
$$
Substituting our expression for $ P $:
$$
P^2 = \left(\frac{AB}{\sqrt{A^2 + B^2}}\right)^2 = \frac{A^2B^2}{A^2 + B^2}
$$
Thus,
$$
\frac{1}{P^2} = \frac{A^2 + B^2}{A^2B^2}
$$
This can be rewritten as:
$$
\frac{1}{P^2} = \frac{1}{A^2} + \frac{1}{B^2}
$$
as required, confirming the relationship between the lengths of the intercepts and the perpendicular from the origin.
Conclusion
This derivation elegantly links the geometric properties of a line in relation to its intercepts and the perpendicular distance from the origin, demonstrating how these elements are interrelated in coordinate geometry.

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