Question

$\huge\red{\fbox{\bf{Question}}}$

If p is the length of the perpendicular from the origin on the line whose intercepts on the axes a and b, then show that
$\frac{1}{p {}^{2} } = \frac{1}{ {a}^{2} } + \frac{1}{ \ {b}^{2} }$
​

Answer 1

$\large\underline{\text{Method}}$

We get the area in a different way, as each triangle contains a pair of perpendicular lines. Let us begin. For the method view the attachment.

$\large\underline{\text{Solution}}$

Generalization.

Let us consider the vertices of the intercepts are $A(a,0)$ and $B(0,b)$.

Triangle area.

$\implies\triangle OAB=\dfrac{1}{2}\times \overline{OA}\times \overline{OB}$

$\implies\triangle OAB=\dfrac{1}{2}\times p\times\overline{AB}$

Length of segments.

The length of $\overline{OA}$

$\implies\overline{OA}=a$

The length of $\overline{OB}$

$\implies\overline{OB}=b$

The length of $\overline{AB}$

$\implies\overline{AB}=\sqrt{a^{2}+b^{2}}$

The area of the triangle.

$\implies \triangle OAB=\dfrac{ab}{2}\text{ (Red triangle)}$

$\implies\triangle OAB=\dfrac{p}{2}\sqrt{a^{2}+b^{2}}\text{ (Blue triangle)}$

Let's compare the area of the red and blue triangle.

$\implies\dfrac{ab}{2}=\dfrac{p}{2}\sqrt{a^{2}+b^{2}}$

Squaring both sides

$\implies\dfrac{a^{2}b^{2}}{4}=\dfrac{p^{2}(a^{2}+b^{2})}{4}$

Thereby,

$\implies a^{2}b^{2}=p^{2}(a^{2}+b^{2})$

$\implies \dfrac{1}{p^{2}}=\dfrac{a^{2}+b^{2}}{a^{2}b^{2}}$

$\implies\dfrac{1}{p^{2}}=\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}$

Hence proven.

Answer 2

Answer:

Step-by-step explanation:

topic :

• traingle of a square and b square

given :

• If p is the length of the perpendicular from the origin on the line whose intercepts on the axes a and b, then show that
• $\frac{1}{p {}^{2} } = \frac{1}{ {a}^{2} } + \frac{1}{ \ {b}^{2} }$

to find :

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

solution :

• the file in attachment please check

answer :

• length of p = 1/p = 1/a + 1/b