If $p$ is the length of the perpendicular from the origin of a line $ \frac{x}{a}+\frac{y}{b}=1 $, what can be it's property?

Question

If $p$ is the length of the perpendicular from the origin of a line $ \frac{x}{a}+\frac{y}{b}=1 $, what can be it's property?

The options were:

A) $ p^2=a^2+b^2 $

B) $ \frac{1}{p^2}=a^2+b^2 $

C) $ p^2 = \frac{1}{a^2} +\frac{1}{b^2} $

D) $ \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} $

I tried to take it to the slope form, and then try the perpendicular, but it is not working... It's providing messy calculations, what am I supposed to do, and how?

Answer 1

Using the standard formula to calculate the perpendicular distance of a line from a point,

$$p^2=\dfrac{(-1)^2}{\left(\dfrac1a\right)^2+\left(\dfrac1b\right)^2}$$

$$\dfrac1{p^2}=?$$

Alternatively,

For $bx+ay-ab=0,$

$$p^2=\dfrac{(-ab)^2}{a^2+b^2}\iff\dfrac1{p^2}=\dfrac{a^2+b^2}{a^2b^2}=?$$

Answer 2

Observe that the line $\dfrac{x}{a} + \dfrac{y}{b} = 1$ passes through two points: $(a,0)$ and $(0,b)$.

Consider the right triangle with vertices $(0,0),(a,0),(0,b)$, then $p$ equals to the height of the hypotenuse.

Now you can find a pair of similar triangles to deduce the value of $p$.

Answer 3

The perpendicular line through the origin is

$$ax-by=0$$ (the slopes of orthogonal lines are opposite inverses of each other).

Hence the intersection point

$$\frac{ab}{a^2+b^2}(b,a),$$ and the squared distance to the origin,

$$\frac{a^2b^2}{a^2+b^2}.$$

Answer 4

$y=0; x=a$; and $x=0, y=b$.

Right triangle with sides $|a|,|b|$ and hypotenuse $c:= \sqrt{a^2+b^2}$.

Area of this triangle:

$(1/2)|ab|=(1/2)pc$, where $p$ is the height on side c (distance from origin to line).

$p^2= \dfrac{a^2b^2}{a^2+b^2}$.