If the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ represent two straight lines,
then prove that the square of the distance of their point of intersection from the origin is $\frac{c (a + b) - f^{2} - g^{2}}{a b - h^{2}}$

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Solution

$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$

Let the lines represented by the given equation be

$l x + m y + n = 0....... (i) l^{'} x + m^{'} y + n^{'} = 0........ (i i) a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = (l x + m y + n) (l^{'} x + m^{'} y + n^{'})$

Equating the coefficients we get,

$l l^{'} = a, m m^{'} = b, n n^{'} = c m n^{'} + m^{'} n = 2 f, n l^{'} + n^{'} l = 2 g, l m^{'} + m l^{'} = 2 h$

Solving $(i)$ and $(i i)$ by cross multipplication

$\frac{x}{m n^{'} - m^{'} n} = \frac{y}{n l^{'} - n^{'} l} = \frac{1}{m l^{'} - l m^{'}}$

So point of intersection P is $(\frac{m n^{'} - m^{'} n}{m l^{'} - l m^{'}}, \frac{n l^{'} - n^{'} l}{m l^{'} - l m^{'}})$

Square of distance from origin $= {(\frac{m n^{'} - m^{'} n}{m l^{'} - l m^{'}})}^{2} + {(\frac{n l^{'} - n^{'} l}{m l^{'} - l m^{'}})}^{2}$

$= \frac{{(m n^{'} - m^{'} n)}^{2} + {(n l^{'} - n^{'} l)}^{2}}{{(m l^{'} - l m^{'})}^{2}} = \frac{{(m n^{'} + m^{'} n)}^{2} - 4 m m^{'} n n^{'} + {(n l^{'} + n^{'} l)}^{2} - 4 n n^{'} l l^{'}}{{(m l^{'} + l m^{'})}^{2} - 4 m m^{'} l l^{'}} = \frac{4 f^{2} - 4 b c + 4 g^{2} - 4 a c}{4 h^{2} - 4 a b} = \frac{c (a + b) - f^{2} - g^{2}}{{a b - h}^{2}}$

Hence proved.

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Similar questions

The square of distance between the point of intersection of the lines represented by the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ and origin, is

Q. If the equation

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

represents a pair of straight lines, then find the square of the distance of their point of intersection from the origin

Q. Statement A: The point of intersection of the lines represented by

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

(\frac{b g - h f}{h^{2} - a b}, \frac{a f - g h}{h^{2} - a b})

Statement B: The point of intersection of

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

(\sqrt{(\frac{f^{2} - b c}{h^{2} - a b})}, \sqrt{(\frac{g^{2} - a c}{h^{2} - a b})})

Q. Prove that the general equation

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

represents two parallel straight lines if

h^{2} = a b

and

b g^{2} = a f^{2}

Prove also that the distance between them is

\frac{2 \sqrt{g^{2} - a c}}{a (a + b)} .

Q. If the equation

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

represents a pair of straigth lines,

then the square of the distances of their point of intersection from the origin

is?

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If the equation ax2+2hxy+by2+2gx+2fy+c=0 represent two straight lines, then prove that the square of the distance of their point of intersection from the origin is c(a+b)−f2−g2ab−h2

If the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ represent two straight lines,
then prove that the square of the distance of their point of intersection from the origin is $\frac{c (a + b) - f^{2} - g^{2}}{a b - h^{2}}$