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Question: If one of the lines of \(m{{y}^{2}}+\left( 1-{{m}^{2}} \right)xy-m{{x}^{2}}=0\) is bisector of the a...

If one of the lines of my2+(1−m2)xy−mx2=0m{{y}^{2}}+\left( 1-{{m}^{2}} \right)xy-m{{x}^{2}}=0 is bisector of the angle between the lines xy=0xy=0, then mm is
(the question has multiple choices as correct option)
A. 11
B. 22
C. −12\dfrac{-1}{2}
D. −1-1

Explanation

Solution

First we will find the line that bisects the angle between the given lines xy=0xy=0. Then we will find the lines that can be formed by the given equation my2+(1−m2)xy−mx2=0m{{y}^{2}}+\left( 1-{{m}^{2}} \right)xy-m{{x}^{2}}=0, from these two we will equate the obtained equation of lines that are obtained from the equation xy=0xy=0. Then we will have a value of mm.

Complete step-by-step solution
Given that,
The pair of equations is my2+(1−m2)xy−mx2=0m{{y}^{2}}+\left( 1-{{m}^{2}} \right)xy-m{{x}^{2}}=0
Simplifying the above equation by expanding the term (1−m2)xy\left( 1-{{m}^{2}} \right)xy, then
my2+(1−m2)xy−mx2=0 ⇒my2+xy−xym2−mx2=0\begin{aligned} & m{{y}^{2}}+\left( 1-{{m}^{2}} \right)xy-m{{x}^{2}}=0 \\\ &\Rightarrow m{{y}^{2}}+xy-xy{{m}^{2}}-m{{x}^{2}}=0 \end{aligned}
Now taking yy common from my2+xym{{y}^{2}}+xy and taking −mx-mx common from the term −xym2−mx2-xy{{m}^{2}}-m{{x}^{2}} in the above equation, then
my2+xy−xym2−mx2=0 ⇒y(my+x)−mx(my+x)=0\begin{aligned} & m{{y}^{2}}+xy-xy{{m}^{2}}-m{{x}^{2}}=0 \\\ &\Rightarrow y\left( my+x \right)-mx\left( my+x \right)=0 \end{aligned}
Now taking the term my+xmy+x as common from the above equation, then
y(my+x)−mx(my+x)=0 ⇒(my+x)(y−mx)=0\begin{aligned} & y\left( my+x \right)-mx\left( my+x \right)=0 \\\ &\Rightarrow \left( my+x \right)\left( y-mx \right)=0 \end{aligned}
Now we will equate the both the terms my+xmy+x and y−mxy-mx to 00 , then the value of mm is
⇒my+x=0 ⇒my=−x ⇒m=−xy...(i)\begin{aligned} &\Rightarrow my+x=0 \\\ &\Rightarrow my=-x \\\ &\Rightarrow m=\dfrac{-x}{y}...\left( \text{i} \right) \end{aligned}
⇒y−mx=0 ⇒y=mx ⇒m=yx....(ii)\begin{aligned} &\Rightarrow y-mx=0 \\\ &\Rightarrow y=mx \\\ &\Rightarrow m=\dfrac{y}{x}....\left( \text{ii} \right) \end{aligned}
Given that one of the above lines is an angular bisectors of the line xy=0xy=0.
The equations of lines required to form the pair of lines xy=0xy=0 are
x=0x=0 , y=0y=0
Now the angular bisectors of the above lines are
x=yx=y , x=−yx=-y
Now we are going to substitute the above two values in the equations (i),(ii)\left( \text{i} \right),\left( \text{ii} \right), then we will have four values for mm.
Substituting x=yx=y in (i)\left( \text{i} \right), then
⇒m=−yy=−1\Rightarrow m=\dfrac{-y}{y}=-1
Substituting x=yx=y in (ii)\left( \text{ii} \right), then
m=yy=1m=\dfrac{y}{y}=1
Substituting x=−yx=-y in (i)\left( \text{i} \right), then
⇒m=−(−y)y=1\Rightarrow m=\dfrac{-\left( -y \right)}{y}=1
Substituting x=−yx=-y in (ii)\left( \text{ii} \right), then
⇒m=−yy=−1\Rightarrow m=\dfrac{-y}{y}=-1
Hence the values of mm are ±1\pm 1.

Note: We can also find the value of mm in simple way that the sum of the slopes of the lines of my2+(1−m2)xy−mx2=0m{{y}^{2}}+\left( 1-{{m}^{2}} \right)xy-m{{x}^{2}}=0 should be equal to 00. The sum of the slopes of the lines are obtained from their pair of linear equation by using the below formula
Sum of the Slopes=−coefficient of xycoefficient of y2\text{Sum of the Slopes}=-\dfrac{\text{coefficient of }xy}{\text{coefficient of }{{\text{y}}^{2}}}
Then the sum of the slopes is equal to 00, then
−coefficient of xycoefficient of y2=0 ⇒−(1−m2)m=0 ⇒1−m2=0 ⇒m2=1 ⇒m=±1\begin{aligned} & -\dfrac{\text{coefficient of }xy}{\text{coefficient of }{{\text{y}}^{2}}}=0 \\\ &\Rightarrow -\dfrac{\left( 1-{{m}^{2}} \right)}{m}=0 \\\ &\Rightarrow 1-{{m}^{2}}=0 \\\ &\Rightarrow {{m}^{2}}=1 \\\ &\Rightarrow m=\pm 1 \end{aligned}
From the both methods we got the same values for mm.