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If the equation $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y=0$ has one line as the bisector of the angle between co-ordinate axes, then
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Verified Answer
The correct answer is:
$(a+b)^{2}=4 h^{2}$
(A)
In the given pair of lines, one line is $y=\pm x \Rightarrow x \pm y=0$
Let the other line be $a x+b y+c=0$
$\therefore(a x+b y+c)(x+y)=0$ or $(a x+b y+c)(x-y)=0$
$a x^{2}+b x y+c x+a x y+b y^{2}+c y=0$ or $a x^{2}+b x y+c x-a x y-b y^{2}-c y=0$
$a x^{2}+(a+b) x y+b y^{2}+c x+c y=0 \quad \ldots(1)$ or $a x^{2}+(b-a) x y-b y^{2}+c x-c y=0 \ldots(2)$
Given eq. is $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y=0 \quad \ldots$ (3)
Eq. (1) and (3) as well as eq. (2) and (3) represent the same line.
Comparing, we write
$2 h=a+b$ or $2 h=b-a \Rightarrow 4 h^{2}=(a+b)^{2}$ among options given.
In the given pair of lines, one line is $y=\pm x \Rightarrow x \pm y=0$
Let the other line be $a x+b y+c=0$
$\therefore(a x+b y+c)(x+y)=0$ or $(a x+b y+c)(x-y)=0$
$a x^{2}+b x y+c x+a x y+b y^{2}+c y=0$ or $a x^{2}+b x y+c x-a x y-b y^{2}-c y=0$
$a x^{2}+(a+b) x y+b y^{2}+c x+c y=0 \quad \ldots(1)$ or $a x^{2}+(b-a) x y-b y^{2}+c x-c y=0 \ldots(2)$
Given eq. is $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y=0 \quad \ldots$ (3)
Eq. (1) and (3) as well as eq. (2) and (3) represent the same line.
Comparing, we write
$2 h=a+b$ or $2 h=b-a \Rightarrow 4 h^{2}=(a+b)^{2}$ among options given.
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