
The equation \[{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0~\] represents
A. A circle
B. Two lines
C. Two parallel lines
D. Two mutually perpendicular lines
Answer
162.9k+ views
Hint: By simplifying the equation \[{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0\] and comparing it with the general form of a circle, straight line we can determine what the equation actually represent. The simplifying value of $x,y$ can also tell us about the equation.
Formula used:
$\tan \theta =\dfrac{2\sqrt{{{h}^{2}}-ab}}{a+b}$
$a{{x}^{2}}+2hxy+b{{y}^{2}}$
Complete step by step solution: The given equation is \[{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0~\].
Simplifying the above equation we get-
\[
{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0~ \\
{{x}^{2}}+2xy+{{y}^{2}}-{{x}^{2}}-{{y}^{2}}=0 \\
2xy=0 \\
xy=0 \\
x=0,y=0 \\
\]
Thus simplifying the above equation we get the value of $x,y$ as $x=0,y=0$ .
So, the given equation represents the pair of straight lines passing through the $x$ axis and $y$ axis.
Now the general formula of a pair of straight lines is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}$. Here $a,~b,~h$ are coefficients.
Now comparing it with the equation we get by simplifying the equation \[{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0\] that is comparing with $2xy=0$ we get the values of $a=0,~b=0,~h=1$.
Now the angle between the straight line is given as follows-
$\tan \theta =\dfrac{2\sqrt{{{h}^{2}}-ab}}{a+b}$
Putting the values of $a,~b,~h$ we get-
$
\tan \theta =\dfrac{2\sqrt{{{h}^{2}}-ab}}{a+b} \\
\tan \theta =\dfrac{2\sqrt{{{1}^{2}}-0}}{0} \\
\tan \theta =\infty \\
\theta ={{90}^{o}} \\
$
Thus the angle between the pair of straight lines is ${{90}^{o}}$ . So, the straight lines are perpendicular to each other.
Thus we can write that the equation \[{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0~\] represents two mutually perpendicular lines.
Thus, Option (D) is correct.
Note: The other method to solve this equation is to determine the sum of the coefficients $a,~b$ . If the sum of $a,~b$ is zero then the straight lines are perpendicular to each other.
Formula used:
$\tan \theta =\dfrac{2\sqrt{{{h}^{2}}-ab}}{a+b}$
$a{{x}^{2}}+2hxy+b{{y}^{2}}$
Complete step by step solution: The given equation is \[{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0~\].
Simplifying the above equation we get-
\[
{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0~ \\
{{x}^{2}}+2xy+{{y}^{2}}-{{x}^{2}}-{{y}^{2}}=0 \\
2xy=0 \\
xy=0 \\
x=0,y=0 \\
\]
Thus simplifying the above equation we get the value of $x,y$ as $x=0,y=0$ .
So, the given equation represents the pair of straight lines passing through the $x$ axis and $y$ axis.
Now the general formula of a pair of straight lines is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}$. Here $a,~b,~h$ are coefficients.
Now comparing it with the equation we get by simplifying the equation \[{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0\] that is comparing with $2xy=0$ we get the values of $a=0,~b=0,~h=1$.
Now the angle between the straight line is given as follows-
$\tan \theta =\dfrac{2\sqrt{{{h}^{2}}-ab}}{a+b}$
Putting the values of $a,~b,~h$ we get-
$
\tan \theta =\dfrac{2\sqrt{{{h}^{2}}-ab}}{a+b} \\
\tan \theta =\dfrac{2\sqrt{{{1}^{2}}-0}}{0} \\
\tan \theta =\infty \\
\theta ={{90}^{o}} \\
$
Thus the angle between the pair of straight lines is ${{90}^{o}}$ . So, the straight lines are perpendicular to each other.
Thus we can write that the equation \[{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)=0~\] represents two mutually perpendicular lines.
Thus, Option (D) is correct.
Note: The other method to solve this equation is to determine the sum of the coefficients $a,~b$ . If the sum of $a,~b$ is zero then the straight lines are perpendicular to each other.
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