If the perpendicular drawn from the origin on $a x + b y + a + b = 0$ is P, then prove that $p^{2} = 1 + \frac{2 a b}{a^{2} + b^{2}}$ .

Open in App

Solution

$x - i n t e r c e p t (A O) : x_{i} = - \frac{(a + b)}{a}$

$y - i n t e r c e p t (B O) : y_{i} = - \frac{(a + b)}{b}$

$A B^{2} = A O^{2} + B O^{2}$

$\Rightarrow A B^{2} = (a + b)^{2} (\frac{1}{a^{2}} + \frac{1}{b^{2}})$

Area of $Δ (A O B)$ taking $A O$ as base and $B O$ as height :

$A = \frac{1}{2} \times \frac{(a + b)^{2}}{a b}$

Area of $Δ (A O B)$ taking $A B$ as base and $p$ as height :

$A = \frac{1}{2} \times A B \times p$

Equating both expressions,

$\frac{(a + b)^{2}}{a b} = A B \times p$

$\Rightarrow \frac{(a + b)^{4}}{a^{2} b^{2}} = (a + b)^{2} (\frac{1}{a^{2}} + \frac{1}{b^{2}}) \times p^{2}$ (squaring both sides)

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

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

Suggest Corrections

Similar questions

Q. If the length of perpendicular drawn from origin to

\frac{x}{2 a} + \frac{y}{2 b} = 4

p

then prove that

\frac{64}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

Q. If length of perpendicular from origin to line

\frac{x}{a} + \frac{y}{b} = 1

is p, then prove that

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

Q. If P is the length of perpendicular from the origin to 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}}

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

Q. If

2 p

is the length of perpendicular from the origin to the line whose intercepts on the axes are

2 a

and

b

, then show that

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

Join BYJU'S Learning Program

If the perpendicular drawn from the origin on ax+by+a+b=0 is P, then prove that p2=1+2aba2+b2.

If the perpendicular drawn from the origin on $a x + b y + a + b = 0$ is P, then prove that $p^{2} = 1 + \frac{2 a b}{a^{2} + b^{2}}$ .