Question

If 2p is the length of the perpendicular from the origin to the line $\dfrac{x}{a}+\dfrac{y}{b} = 1$, then ${a^2}$, $8{p^2}$ and ${b^2}$ are in
1). AP
2). GP
3). HP
4). None of these

Answer 1

Hint: First we will find the distance between the line and the point using the distance formula. Then, we will arrange the equation formed in the way we will get the answer. After solving step by step we will come to know whether they are in AP, GP or HP.

Complete step-by-step solution:
Given:
Line = $\dfrac{x}{a} + \dfrac{y}{b} = 1$
Distance (d) = 2p
Distance of the line is from the origin. So, point $(x_1, y_1)$ is equal to (0, 0)
Now we will use distance formula to make the equation which we need to find the answer:
$d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$
Now, putting the values in the above formula we get,
$2p = \left| {\dfrac{{0 + 0 + \left( { - 1} \right)}}{{\sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} }}} \right|$
By squaring on both sides. We get,
$4{p^2} = \dfrac{1}{{\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}}}$
Now, we will reciprocal this equation
$\dfrac{1}{{4{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}$
We want 8p2 in the equation to find the answer. So, we will multiply and divide L.H.S. with 2
$\dfrac{2}{{8{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}$
This suggests that $a^2$, $8{p}^2$ and $b^2$ are in HP.
So, option (3) is the correct option.

Note: A harmonic progression (HP) is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic progression (AP) that does not contain 0. There is one more formula of calculating distance which is $\left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|$but this is for calculating the perpendicular distance of a point from a 3-D plane not a line. So, don’t get confused with it.