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/p²=1/a²+1/b².

Answer 1

The relationship
`\( (1)/(p^2) = (1)/(a^2) + (1)/(b^2) \)` holds true, where p is the perpendicular distance from the origin to a line with intercepts a and b on the axes.

Let the line have intercepts a and b on the x- and y-axes, respectively. The coordinates of the points where the line intersects the axes are (a, 0, 0) and (0, b, 0).

The vector representing the line is given by
`\(\vec{v} = (a, b, 0)\)`. Now, we need a vector from the origin to any point on the line, say
`\((a, b, 0)\)`, which is
`\(\vec{r} = (a, b, 0)\)`.

The length of the perpendicular from the origin to the line is given by the projection of
`\(\vec{r}\)` onto
`\(\vec{v}\)`. The length p is given by:

`\[ p = \frac{\vec{r} \cdot \vec{v}}{\|\vec{v}\|} \]\[ p = ((a, b, 0) \cdot (a, b, 0))/(√(a^2 + b^2)) \]\[ p = (a^2 + b^2)/(√(a^2 + b^2)) \]`

Now, let's simplify
`\(1/p^2\)`:

`\[ (1)/(p^2) = (1)/(\left((a^2 + b^2)/(√(a^2 + b^2))\right)^2) \]\[ (1)/(p^2) = (1)/(a^2 + b^2) \]`

Now, we need to show that
`\( (1)/(p^2) = (1)/(a^2) + (1)/(b^2) \)`:

`\[ (1)/(a^2 + b^2) = (1)/(a^2) + (1)/(b^2) \]\[ (1)/(a^2 + b^2) = (b^2 + a^2)/(a^2b^2) \]\[ (1)/(a^2 + b^2) = (1)/(a^2b^2) \]`

This is indeed equal to
`\( (1)/(p^2) \)`, thus proving the given relation
`\( (1)/(p^2) = (1)/(a^2) + (1)/(b^2) \)`.